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Power Factor Theory---------------------------------------Aradhana RayConsulting Engineer Doble Engineering, India

Basic Insulation & Power Factor Theory

The underlying principle of the Doble M4000 Test isto measure

the fundamental AC electrical characteristics of insulation.

•Changes in the electrical characteristics of insulation can indicate:

- An increased or decreased size of the insulation system,- Presence or absence of an insulation component or the movement of the conductors. -Indicate presence of moisture, insulation deterioration, destructive agents or ionization

- These changes can effect the performance of the insulation system.

Basic Insulation & Power Factor Theory Insulation

IEEE Defines Insulation as:

“Material or a combination of suitable non-conducting material that provides electrical isolation of two parts at different voltages.”

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Dielectric

implies that the medium or material has specific measurable properties, such as:

Dielectric Strength, Dielectric Absorption, Dielectric Constant, Dielectric Loss & Power Factor

Examples of Material With Insulating Properties

Gaseous Liquid Solid

High Vacuum Hydrocarbon- Based Oil

Cellulose

Air Silicone Oil Porcelain

Sulfur Hexafluoride (SF6)

Distilled Water Phenolics

Insulation

Insulation is basically two plates separated by one or more dielectrics. One plate is at a high potential and the other at a lower or ground potential.

dielectric insulationCurrent generated by polar contaminants in the

dielectric shows up as Watts.

Heat/Watts

Fundamental AC Electrical Characteristics

•Dielectric-Loss•Power Factor

•Capacitance

•Power Factor Tip-up

•Total Charging Current

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Total Charging Current

Typical Insulation System

Good Insulation:Has a very lowpower factor CP RP

• IR<<IC for most insulationsystems, IC ~ IT

IT

IC IR ~ 0

The capacitance current and the resistive current cannot be just arithmetically added together because the quantities vary in time and are not in phase.

IITT = I= ICC + I+ IRR→→ →→ →→ Voltage and Currents

-1.5

-1

-0.5

0

0.5

1

1.5

0 90 180 270 360 450 540 630

Angle

Mag

nitu

de

VoltageIRICIT

0.750.75

0.250.25

0.790.79

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Capacitance

Parallel Circuit

CP

E

IT

IRIC

Basic Circuit

The Perfect Capacitor

The perfect capacitor

passes no Direct CurrentAlternating Current leads the

voltage by 90 degreeshas a Power Factor of 0% by

definitionPerfect

Capacitor% PF = 0%

Capacitor Current

-1.5

-1

-0.5

0

0.5

1

1.5

0 90 180 270 360 450 540 630

Angle

Mag

nitu

de Voltage

Current

5

E

IC

θ=90ο

Alternating capacitance current leads the voltage by 90 degrees

Capacitor Current Vector Diagram

The Capacitor

Ad

Two conducting plates with area A separated by a dielectric with a thickness of d and dielectric constant ε

Dielectric

Plates

Capacitance

C = Capacitanceε = dielectric constant d = Distance between plates

•All of these variables are Physical Parameters

AdC= Aε

4πd

Dielectric Constant

In 1836, Michael Faraday discovered that when the plates between a capacitor were filled with another insulating material, the capacitance would change.

This factor is the dielectric constant εBy definition the dielectric constant of a Vacuum is 1.0. All other

dielectric constants are referenced to this standard.

Vacuum

Cvacuum=10 pF

Oil ε=2.2

Coil = ε x CVacuum = 22 pF

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Material Dielectric Constant Dielectric Strength (kV/mm)Vacuum 1.0Air 1.000549 3Mica 5.4 10Oil Impregnated Paper 3.7 16Porcelain 7 5.7Rubber 2.4 - 3.7 12Oil 2.24 12Silicone Fluid 2.75 15

Various Dielectric Constants

Test Mode-GST Ground

Test Ground

Test-Set Ground Lead

High-Voltage Test Cable

Oil = 2.1∈Porcelain = 7.0∈Paper = 2.0∈

Guard

Current & Loss Meter

I

Typical Insulation System

Measuring the Dielectric Constant of a Material

IIOilOilOilOil

IOil COil

CVac== εoil = 2.24

IVac

An alternating current of the same voltage is applied to the capacitor for both tests.

E∼VacuumIIVacVac

E∼

Example: Oil leaking from an Insulation System

Oil = 2.1Porcelain = 7.0Paper = 2.0

Given three dielectrics in series the dielectric constant ε is:

εbefore = 2.1 x 7.0 x 2.0

2.1 + 7.0 + 2.0= 2.65

If the Oil leaks out and is replaced by air...

Air = 1.0

εafter = 1 x 7.0 x 2.0

1 + 7.0 + 2.0= 1.4

C => It

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IC = EωC (ω = 2πf)

} Physical}ConsideredConstantsDuring Testing10 kV, 50 Hz

Ad

IC

Capacitive current and Capacitance

C = Aε4πd

IC = EωAε

4πd

d

ε (Dielectric constant) & d (Distance between plates) are constant.

Ad

A

2C =

Capacitance Change due to Change in Area of Plate

Area of the plates doubles.

(2A)ε4πd

C = Aε4πd

Area of the Capacitor

Area ↑ C

Area ↓ C

↑ Ic ↑

↓ Ic ↓

IIdd

E ∼ d

I2d

E ∼2d

C = Aε4πd

I2d = Eω(C/2)

Distance Between the Plates “d” of the Capacitor

Id = EωC

Aε4π(2d)

= C2

Double the distance

= Id/2

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Distance Between the Plates “d” of the Capacitor

d ↑ C

d ↓ C

↓ Ic ↓

↑ Ic ↑

Insulation System 1995 New 1996 Fault DifferenceCH pF 2155 2159 0.2%CHL pF 4360 3886 -10.9%CL pF 8291 9339 12.6%

Example of Winding Movement

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A ↑ C ↑ Ic ↑A ↓ C ↓ Ic ↓

Summary of Possible Capacitance Changes of a Capacitor

ε ↑ C ↑ Ic ↑ε ↓ C ↓ Ic ↓d ↑ C ↓ Ic ↓d ↓ C ↑ Ic ↑

Changes in Current/Capacitance

SignificanceIndicate a physical change

Bushings - shorting of capacitive layersTransformers - movement of core/coilsArresters - broken elements

Suggested Limits+ 5% - Investigate+ 10% - Investigate/remove from service

Power Factor

Basic Power Factor

E CP RP

IT

ICIR

IT=Total Current IC=Capacitive CurrentIR=Resistive Current E=Applied Voltage

ITIC

IR E

O

0% PF

100% PF

θ =Power Factor Angle

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ITIC

IR E

O

E CP RP

IT

ICIR

P.F. = WattsE × IT

P.F. = IRIT

P.F. = cosθ

Power = Voltage x Current x Cos(θ)

P= E IT Cos(θ) What Is Power Factor (PF)?

Watts = Cosine E IT

× × θ

PF = Cosine = WattsθE I

T×

=××

=E IE I

II

R

T

R

T

Watts = E x IR

What Is Power Factor (PF)?

=WmAX10

Power Factor WI E

al PowerApparent PowerT

= =*Re

To express power factor in percent (% PF), multiply by 100:

%*

PF WmA

= −X XX

10 3 10 103 100

10 kV equivalent values

Limits of % P.F.

CapacitorPF = 0%

ResistorPF = 100%

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θ (°) P. F. (%)90 0 (Capacitance only)89.714 0.589.427 1.045 70.730 86.60 100 (Resistance only)

Power Factor Relationships

ITIC

IR

O

ITIC

IR

O

IIL

O

EE

IT

EE

O

IT

Increase in LossesIncrease in Losses Added InductanceAdded Inductance

Changes In Power FactorChanges In Power Factor

Capacitors, Resistors, and Inductors

Basic Theory

0 90 180 270 360

IR

↓↓ ↓ILIC

E↓

Power Factor Is Size IndependentIC2

Specimen 1: 5 MVA TransformerSpecimen 2: 10MVA Transformer

remains the same regardless of the size of the transformer

Power Factor is an evaluation of the quality of the insulationand is size independent

IT2

IR2 EIR1

IC1 IT1

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Definition of Power Factor

Power factor is a measurement of the efficiency of insulation

system.

• Dielectric loss is a function of volume. For a larger insulation system, there is more material to dissipate watts due to inherent losses, deterioration, and contamination.

• To analyze losses there is a need to be able to compare the size of the insulation tested, which is difficult to measure physically.

Why Power Factor Instead of Dielectric Loss, Watts?

• Provides an index to compare the relative losses of different sizes of insulation systems.

• The power factor is the ratio of the real power to the apparent power or the resistor current to the total current. PReal /PApparent= IR/IT

• Minimizing the power factor will provide an insulation system that is in better condition.

• A lower power factor insulation system will have lower relative losses.

Power Factor

IC IT

IR EE

IC IT

IREE

2IC 2IT

2IRIIRR/I/ITT = (I= (IRR+I+IRR)/(I)/(ITT+I+ITT) = 2I) = 2IRR/2I/2ITT = I= IRR/I/IT T = PF= PF

Power Factor Relationships

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Voltage and Currents

-1.5

-1

-0.5

0

0.5

1

1.5

0 90 180 270 360 450 540 630

Angle (360 degrees = 1 / 60th second)

Mag

nitu

de

IC IT

IR EE

Time and Angle RelationshipsVoltage and Currents

-1.5

-1

-0.5

0

0.5

1

1.5

0 90 180 270 360 450 540 630

Angle

Mag

nitu

de

VoltageIRICIT

%PF = 0%PF = 0

Voltage and Currents

-1.5

-1

-0.5

0

0.5

1

1.5

0 90 180 270 360 450 540 630

Angle

Mag

nitu

de

VoltageIRICIT

%PF = 0.5%PF = 0.5

Voltage and Currents

-1.5

-1

-0.5

0

0.5

1

1.5

0 90 180 270 360 450 540 630

Angle

Mag

nitu

de

VoltageIRICIT

%PF = 1.0%PF = 1.0

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Voltage and Currents

-1.5

-1

-0.5

0

0.5

1

1.5

0 90 180 270 360 450 540 630

Angle

Mag

nitu

de

VoltageIRICIT

%PF = 50%PF = 50

Voltage and Currents

-1.2-1

-0.8-0.6-0.4-0.2

00.20.40.60.8

1

-90 -60 -30 0 30

Angle

Mag

nitu

de

VoltageIRICIT

PowerPowerFactorFactorAngleAngle

Dielectric Loss and Power Factor:

The Dielectric Loss and Power Factor are sensitive to soluble polar,ionic or colloidal materials:

Moisture (free, in cellulose, with particles in oil)

Products of Oxidation or mineral oil

Carbon (with moisture)

Metal Soaps

At Higher voltage:

Ionization in solid insulation

Basic Principals of Testing:Always test the smallest piece possible

TestInsulation

contamination

• Power factor testing measure the average condition of an insulation system

• Contamination would affect the total insulation system, but not to a large degree

Always break an insulation system into the Smallest possible part in order to detect insulation faults.

TestThe contamination becomes a “bigger piece” of

the insulation and is easier to see

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P F∑

= =. . .5 5 5 + + + .54

.5

1 12

12

12

12

2c

= + + + = c p F=12

PF = .5

2 pF

PF = .5 PF = .5 PF = .5

2 pF 2 pF 2 pF

PF x∑

= + =. . .5 3 2 54

1 0 c pF= 12

PF = .5

2 pF

PF = .5 PF =2.5 PF = .5

2 pF 2 pF 2 pF

1)

Is the Doble Test Effective for Detecting Defective Insulation?

PF = .5

2 pF

PF = .5 PF = .5

2 pF 2 pF

PF =+ +

=. . . .5 5 5

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1 12

12

12

15c

= + + = . c =.667

2)

Is the Doble Test Effective for Detecting Defective Insulation? (continued)

Analysis of Percent Power Factor

Compare to limits published by DobleCompare to previous resultsCompare among similar or sister unitsExamine the trendDo not use PF if current is less than 300µA

Power Factor vs. Dissipation Factor

E

Θ

δ

IR

IC IT

Power Factor = = II

Dissipation Factor = = II

R

T

R

C

COS

TAN

Θ

δ

Θ° % PF (% COS Θ) δ° % DF (% TAN ∆)90 0 0 0

89.71 .500 .29 .50084.26 10.00 5.74 10.05

0 100.00 90 INFINITY

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= II

R

T

orPower Fact = cos θ

R

C

= II

Dissipation Factor = tan δ

90 0 0 0θθ ° % PF (% COS ) δ° % DF (% TAN ∆)

89.71 .500 .29 .50084.26 10.00 5.74 10.05

0 100.00 90 INFINITY

Power Factor Vs. Dissipation Factor

E

θ

δ

IR

IC IT

θ % PF (% cos) δ % DF (% tan)90 0 0 0

89.71 .50 .29 .5087.13 5.00 2.87 5.0084.26 10.00 5.74 10.0581.37 15.00 8.63 15.1853.13 60.00 36.87 75.0045.00 70.71 45.00 100.00

0 100 90 infinity

Comparison of Percent Power Factor With Percent Dissipation Factor for Various Phase Angles of Θ and δ

Power Factor Tip-Up

Voids and the Power Factor Tip-Up TestWhen we closely examine insulation, very small gaps or “voids” exist. These voids develop an electrostatic potential on their surfaces. These small gaps become ionized: Partial Discharge/Corona.

Voids

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Partial Discharge (Corona)

•The phenomenon of an electrical discharge that does not completely bridge the insulation between electrodes or conductors.

Corona -- accompanied by a faint glowPartial Discharge -- may not be luminous (preferred term)

•Partial discharge occurs in:

A void within an insulation system where the voltage gradient is sufficiently high -- has a damaging effect on surrounding materials.

Power Factor Vs. Test Voltage

As test voltage is increased, the power factor will increase dependingon the void density.

Tip-Up = Power Factor at Line-to-ground voltage -Power Factor at 25% Line-to-ground voltage

25% L-G L-GE

%PF

%PF @ 25% L-G%PF @ L-G

Tip-up occurs in dry-type insulation specimens such as Dry Type Transformer, generators, etc.….

Thank You.

Any questions?

aray@doble.com