Improving Electrical Installation Power Factor

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s Merlin Gerin s Square D s Telemecanique

WHAT YOU HAVETO REM EM BEREconomic considerations, directly

linked to local billing practices,

encourage investments with the

aim of improving the power factor

of an electrical installation.

The use of capacitors on networks

with harmonics can have

an amplifying effect. I t is therefore

necessary to add filtering

components to the capacitor bank.

It is important to call in a specialist

to determine the compensation

bank characteristics.

Improving the power factorof an electrical installation

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REM INDERS THE CAPACITOR WITH A DIRECT VOLTAGE

A capacitor can be represented by an insulating layer between two conductorplates (armatures).

If a direct voltage is applied across the terminals the armatures are charged

with a quantity Q of electricity.

Energy W is stored and an electrical field E is established in the dielectric.

A s soon as the capacitor is charged, the current stops flowing ( … except for

a very small quantity of leakage current).

E = -V/e Q = C • V

W = V • Q = C • V2

C = ε • S/e ε • = ε0 • εr

C : capacitor capacitance in Farad (F)ε • = dielectric permittivity in F/m

εr = relative permittivity of the insulating material

ε0 = 8.85 • 10-12 F/m

The unit for measuring capacitance in the M K SA system is the Farad.

Since this is a very large value, the sub-units are typically used (microfarad).

THE CAPACITOR WITH AN ALTERNATING VOLTAGE

Energy is stored and restored 100 or 200 times per second depending on

the network frequency. A current flows, rather than transitory it is periodic,

corresponding to the capacitor charge and discharge.

v = V0 sin ωt

i = dQ /dt et Q = C • V

i = C • dV/dt = Cω V0 cos ωt

As an rms valueI = C ω V

i leads v by 1/4 of a cycle

Improving the power factor of an electrical installation

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+ -

V=

+ -

+ -

+ -

+ -

+ -

+ -

+ -

+ -

+ -

E

e

I

I

V

t

V

2ϕ = -

π

ωt

Advanced quadrature current I = C ω V

ϕ is the phase displacement of the voltage relative to the current



REM INDERS (cont’d) INDUCTANCE (SELF-INDUCTANCE) WITH A DIRECT VOLTAGE

The passing of current through a turn creates a magnetic flux density Bin Tesla (T).

The coil stores the magnetic energy.

The variation of B creates an f.c.e.m. which opposes the passing of the current.

The voltage at the terminals of the reactor is: v = L • di/dt

L = coil inductance in Henry (H ).

INDUCTANCE (SELF-INDUCTANCE) WITH AN ALTERNATING VOLTAGE

The magnetic flux density B is alternating.

i = I0 • sin ωt

v = L • di/dt = Lω I0 cos ωt

As an rms valueV = L ω I

v leads i by 1/4 of the cycle


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+ -

V

I

B

=

I

IV

t

V

2ϕ = + π

ωt

C

LR

Z

Delayed quadrature current V = L ω I

IMPEDANCE

z = R + jX with jX = jLω + (- j 1/Cω)

X = Lω - 1/Cω

equal to



PHASE DISPLACEMENT OF THE CURRENT RELATIVE TO THE VOLTAGE


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REM INDERS (cont’d)

IRI

UjXI

ϕ

P

SQ

ϕ

u = Z • i

ϕ phase displacement of the current relative to the voltagetgϕ = X/R

u(t) = U o sin ωt i( t) = U o/IZIsin (ωt - j)

APPARENT POWER

This is the product of the voltage and the current.

For single phase S = UIFor three-phase S = √

—

3 UI expressed in kVA

ACTIVE POWER

This is the product of the voltage, the current and the cos ϕ

for single phase P = UI cos ϕ

for three-phase P = √ —

3 UI cos ϕ expressed in kW

The active power is transformed into mechanical and thermal power.

Comment: the power of a motor expresses the mechanical power

available through the drive shaft.

The active power takes account of the motor’s efficiency η:

P = Pm/η



REM INDERS (cont’d) REACTIVE POWER

This is the product of the voltage, the current and the sin ϕ

for single phase Q = UI sin ϕ

for three-phase Q = √ —

3 UI sin ϕ expressed in kVAR

The reactive power is transformed into magnetizing power.

The reactive energy consumed is measured by the value of:

tg ϕ = reactive power/active power = Q (kVA R )/P (kW)

For M V metering, tg ϕ = Q /P; for LV metering, it is necessary to add

the transformer losses (in general, the chosen value is tg ϕ’ = tg ϕ + 0.09).

ENERGY FLOW IN THE NETWORKS


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LossesMainly joule

Reactive energy

Generation

Consumers

Public and privatedistribution networks

R

G

Active energy

Three major functions are involved in the electrical networks:

generation, transmission and distribution, and consumption.

Active energy is produced by generators, transmitted and distributed

to the consumers who use it either in mechanical form (motor), in thermal

form (heating), or in chemical form (electrolysis).

The current produced from line and equipment heating: these are the Joule

losses.

Reactive energy is permanently exchanged in the networks between the

reactive power generators (capacitors, synchronous compensator, alternators

under certain conditions) and equipment with magnetic circuits.

Reactive energy in the process of being exchanged is not a loss;but it can cause losses!

c increase in voltage drops, etc.

c

increase in Joule losses (the current transmitted is even higher)c and above all financial losses according to the billing used in the country.



COM PENSATION WHY COM PENSATE?

In designing our electrical installation, we can note the following energyexchanges:


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Losses

Active energyReactive energyProduction

Consumption

Consumption

Losses


Optimization of energyproduction management

Losses


A local reactive energy sourceavoids transmission of this energy

Consumption

U se of a bank with shunt compensation as near as possible to the consumer

enables the reduction of the reactive power which is transmitted on the lines

and thereby reduces losses.

Compensation of energy is intended to limit line voltage

drops by reducing the Joule losses due to transmission

of reactive energy (shunt compensation) or by creating

a capacitive voltage drop (series compensation).

A dding a reactor to the capacitors enables filters to be created and limits

harmonic transmission.

C apacitor use near the generators enables optimization of their operation.



COM PENSATION (cont’d) The additional consequences of compensation are therefore to:

c

enable more economical sizing of the installations (lines, wires, transformers,circuit breakers);

c improve the electrical current quality (filtering) ;

c reducing the level of voltage harmonic distortion to an acceptable value for

the consumer and the distributor.

THE PRINCIPLE OF COM PENSATION

The following graph corresponds to a series compensation arrangement.

The analog graph exists in shunt compensation but in terms of current.


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R.I

U

j.Lω.I

ϕ

j.I

Cω-

R.I

U

j.Lω.I

ϕ

j.I

Cω-

R.I

U

j.Lω.I

ϕ = 0

j.I

Cω-

R.IU

j.Lω.I

ϕ

j.I

Cω-

The reactor is a consumer of reactive energy.

The capacitor is a source of reactive energy.

The addition of capacitors enables the reduction of reactive energy which

must be transmitted on the lines.

Partial compensation

GOOD

The reduction of cosϕ is adjusted

as a function of the load by using

multi-step capacitors.

Total compensation

theorically IDEAL

D ifficult to implement especially

if the load varies.

Over-compensation

DANGER

D angerous overvoltages

the equipment.



PUTTING IT IN PRACTICE THE SERIES CAPACITOR

Series compensation enables the addition of a capacitive voltage dropwhich opposes the inductive line voltage drop. Downstream from the capacitor

the line voltage is higher than the voltage upstream.

For a given transmitted power, the current will therefore be lower and the losses

due to the Joule effect reduced.


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CZ ligne

ChargeU2U1

ϕ

R.I

jX.l

U2

U1 com pensa te d U 1 u n

c o m p e n s a t

e d

∆ U u n c o m p e

n s a t e d

∆ U c o m p e n s a t e d

ϕ

j.I

Cω-

I

The total line current passes through the series capacitor bank. This current

can therefore not be very high.

The use of series capacitor banks is generally limited to the very long EHV

lines (more than 500 km).

For a given U 2 voltage, compensated U 1 is lower than uncompensated U 1.

For a given U 1 voltage, compensated U 2 is higher than uncompensated U 2.

A t the end of the line, the total voltage drop is lower.

A t full load, the current is inductive.

With no load, the line is capacitive; it is necessary to regulate compensation

to avoid overcompensation.



PUTTING IT IN PRACTICE

(cont’d)

THE SHUNT CAPACITOR

For utilit ies companies It enables the reactive energy consumed by the customers to be compensated

for and enables the current upstream of the load to be optimized.


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C

Z line

LoadU2

C

Z line

Optimizespurchasingcosts

Limits joule lossesand optimizes

purchasing costsLoad

It is possible to transmit an even higher active power on a same installation.

It is essential to guarantee that in case of the capacitor banks failing,

the protection devices will respond to avoid overloading the installation.

For power consumers

Installed close to the metering unit the shunt capacitor enables use of

the utility company’s billing contract to be optimized.

In addition, installing capacitors across the current consumers’ terminals

enables the losses due to the Joule effect in the consumer’s installation

to be limited.

A dding a shunt capacitor bank enables the current flowing in the lines

to be reduced.

U2ϕ corrected

ϕU2

Z=

I

line currentbeforecompensation

j Cω U2

The current flowing in the line is lower, the Joule losses are therefore reduced.

D epending on how reactive energy is billed by utility companies, energy

costs can be greatly reduced.

Three types of contracts exist:

c payment for reactive energy consumed starting from a cos ϕ threshold

as in France and Italy;

c a bonus-penalty according to the value of the cos ϕ relative to a thresholdvalue as in Spain or in Egypt;

c a kVA contract with a threshold of overbilling over a period of time as

in the U K and in Canada.



PUTTING IT IN PRACTICE

(cont’d)

INSTALLING SHUNT

CAPACITORS

For production plant

We can consider three situations:The main production plant permanently supplies a networkIn this case, i t can be necessary to use a shunt capacitor bank to maintain

the cos ϕ of the installation at the optimal level of cos ϕ for the alternator.

A lternators are guaranteed for operating conditions at a given cosϕ value

(generally 0.8). If the network functions at a lower cos ϕ level, it will be

necessary to raise it and use a shunt capacitor bank.

The production plant can be coupled to a main networkThis is the case with for private production plants of major industries who

supply energy to the local utilities company.

Depending on billing practices, it is vital to make sure that the generator does

not deteriorate the network cos ϕ. A shunt capacitor bank can prove essential.

The generator is a back-up generatorThe main supply has failed and it is necessary to make sure that the main

capacitor banks are adjusted to the load after load shedding. A small correction

of cos ϕ can be considered.

The following diagram summarizes the various installation possibilities.


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HV bank on an HV distribution network

M V bank on an M V distribution network

M V bank for an M V customer

LV bank set or regulated for an LV customer

LV bank for an M V customer

LV bank for individual compensation

LV customer MV customer MV customer



APPLICATION EXAM PLE For an installation with two motors:


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800 kVA400 V 50 Hz 3 phase

2 motors 250 kWcos ϕ = 0.75

The power necessary to supply the two motors is:

S = P/(η x cosϕ) orη = mechanical efficiency of the motor≈ 0.8

S = 2 x 250/(0.8 x 0.75) = 833 kVA

The 800 kVA transformer is overloaded.

If we compensate this installation with a shunt bank connected to the busbars

in order to have a cos ϕ = 0.92

the power equipment becomes:

S' = 2 x 250/(0.8 x 0.92) = 679 kVA

The transformer is thereby relieved and we have a power reserve.

The capacitor bank to install is defined by:

tg ϕ = Q /P if cosϕ1 = 0.75 tg ϕ1 = 0.88

if cosϕ2 = 0.92 tg ϕ2 = 0.43

therefore (tg ϕ1 - tg ϕ2) = 0.88 - 0.43 = 0.45 = (Q 1 Q 2)/P = Q c/P

Q c is the power necessary for the capacitor bank.

Q c = 250 x 2 x 0.45 = 225 kVAR

P

S 2Q

Qc

ϕ2

ϕ1

S 1

In this theoretical example,

we cannot overload the transformer

and therefore can economize the

billing of 225 kVA R reactive energy.

Generally passing from cos ϕ = 0.8 to 0.93:

c enables the reduction of line losses by 30% with

constant active power;c increases the power transmitted or delivered by

a transformer by 20% at constant losses.



HOWEVER

COM PENSATING CAN BEA SOURCE OF TROUBLE

THE M OTOR WHICH BECOMES A GENERATOR

To supply the reactive energy that is needed, a capacitor bank has beeninstalled across the terminals of a motor.

The unit is controlled by a contactor.

When the motor is started, the capacitors charge.

When the motor drives a high inertia load and when, after the voltage supply

is interrupted, it can continue to run using the kinetic energy of the load

(generally in motors which run at high speeds 3000 tr/mn), it can self-excite

its circuit using the discharge current of the capacitors and function as

an asynchronous generator.

This self-excitation can cause overvoltages greater than the maximum voltage

levels for the network.To avoid this problem, it is necessary to verify that the bank power is less than

the power required for excitation.

Ic capacitor current = 0.9 Io no-load motor current

(see the technical data sheet R ectiphase AC 0303/1).

To avoid this problem, the capacitor bank can be connected to the busbars

with its own breaking device.

COMMUNICATING VESSELS: OVERCURRENT DURING CAPACITOR BANK CLOSING


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When a coupling device is closed, the discharge current from one capacitor

into another is only limited by busbar impedance (in LV at the instant of closing,

a resistance is integrated into the circuit for the duration of the current peak).

To avoid a strong overcurrent, it is necessary to insert a reactor between the

capacitors.

This inductance of closing current limitation is negligible comparedwith the compensating action of capacitors (see cahier technique CT142) .

Qc

PnIo

M

QcIo

M



HOWEVER

COM PENSATING CAN BEA SOURCE OF TROUBLE

(cont’d)

TOO MUCH OF A GOOD THING: OVERCOMPENSATING


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c in the case of an overcompensated network under no load,

ϕ is very close to - π/2

R is low compared to X and the voltage drop is close to - X • I

At the end of the line there is no voltage drop but rather an increase in voltage.c in practice, this type of case is encountered when a capacitor bank is left

connected to a low load (absence or poor functioning of the regulation system).

To avoid overcompensating, it is necessary to use multi-step capacitor banks

and to include the protection devices that will be required in case of

malfunctioning.The problem of overcompensating can occur if no precautions

are taken on industrial networks at night.The overvoltage is limited if the

capacitor bank power is lower than 15% of the transformer one .

THE NEUTRAL POINT OF THE BANK IF IT EXISTS, M UST FLOAT

The potential of the neutral point must remain floating to avoid anovervoltage in one of the steps of the capacitor bank in case of line failure

or ferro-resonance. Generally, capacitor banks are wired:

R.I

jX.l

U2

U1π2

ϕ = -

I

Double star if: network U > 11 kV and Q c > 600 kVARnetwork U < 11 kV and Q c < 1200 kVA R

D elta if: network U ≤ 11 kV and Q c ≤ 1200 kVAR

In MV, in the double star arrangement, the circulating of current betweentwo neutral points is constantly monitored to detect internal capacitor faults.

U se of capacitor banks enables an installation to be sized economically.

Nevertheless, the installation must be protected in order to be able to function

in complete safety without capacitors. The network must be studied and

protected in an adequate manner.



DEALING WITH

HARMONIC POLLUTION(see cahier technique 152)

Electrical power is generally distributed in the form of three voltages making

up a three-phase sinusoidal system. O ne of the parameters of this system is

the wave form which must be as near as possible to sinusoidal.

In networks with harmonic current and voltage sources, such as arc furnaces,

static power converters, lighting, etc., it is necessary to correct the wave if

it exceeds certain deformation limits.

MAIN DISTURBANCES (see cahier technique 152)

Harmonic voltages and currents superposed on the fundamental wave

conjugate their effects on the devices and equipment used.

These harmonic values have varying effects according to the current

consumers encountered:c either instantaneous effectsVibrations, noise, disturbance on low current lines (telephone, monitoring

and control line), disturbances electronics systems, metering systems, etc.);

c or effects over time due to heating.

HARMONIC GENERATORS

In the industrial range, we can give as examples:

c static converters;

c arc furnaces;

c lighting;

c saturated reactors;

c others such as rotary machine gears and

harmonics which are often negligible.

THE QUALITY OF ENERGY

In the absence of capacitor banks, the harmonic pollution is generally limited

and proportional to the polluters’ currents.

Harmonic currents flow along the network and cross the transformers.

The presence of a capacitor bank causes parallel resonance that can lead

to dangerous harmonic pollution.

R esonance exists between the capacitor bank and network reactance

across the bank terminals.

If the order of the resonance is the same as that of the currents injected by

the polluter, the result is an amplification which is absorbed to a greater of

lesser extent by the harmonic values (currents and voltages). This pollution

can be very dangerous for equipment.


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DEALING WITH

HARMONIC POLLUTION(cont’d)

(see cahier technique 152)


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F7 F11F5

Harmonic filtering is a technique based on reactors and capacitors tuned

to the frequencies to be eliminated.

The calculation of harmonic filters must be performed by specialists.

network

Harmonicgenerator

Other currentconsumers

To avoid resonance being dangerous, it is essential to place it outside

of the injected spectrum and/or to absorb it.

This is achieved by increasing the series inductance with a capacitor bank.

HARMONIC FILTERS

By themselves, capacitors do not create harmonics.

They are capable of absorbing high frequency currents

without distorting the voltage (U = I/Cω).

By their presence, the capacitors can reduce the harmonics

already present in the network.



CHOICE OF BREAKING

DEVICES

For example, we can mention the case of an aluminum factory which before

filtering had a voltage distortion level of 3.78% with a cos ϕ = 0.75 and after

compensation and filtering a level of distortion of 0.87% and a cos ϕ = 0.92


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F74.403

F113.66

F55.487

Total Q =13.55 MVARQ to calculate

Filters

20 MWcos ϕ 1 = 0.75cos ϕ 2 > 0.9

Sn = 50 MVAScc = 330 MVA

Scc = 200 MVA

Distortion Ih /I1%

Harmonic number

Without filtering

With filtering

Gh = 40 MVA

11 kV

63 kV

1.1

1

0%

10%

20%

30%

40%

1 3 5 7 11 13 17 19

P

DEALING WITH

HARMONIC POLLUTION(cont’d)

(see cahier technique 152)

C ontrol of capacitor banks requires the use of a breaking device in which the

nominal current is derated in order to avoid heating by harmonics.

G enerally, derating is to the order of 30% .

(see M T partenaire, chapter B-3-2).

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