Improving the power factor
of an electrical installation
WHAT YOU HAVE
TO REMEMBER
Economic 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. It 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.
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■ Merlin Gerin ■ Square D ■ Telemecanique
Improving the power factor of an electrical installation
THE CAPACITOR WITH A DIRECT VOLTAGE
REMINDERS
+
A capacitor can be represented by an insulating layer between two conductor
plates (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.
As soon as the capacitor is charged, the current stops flowing (…except for
a very small quantity of leakage current).
=
V
+
+ E
+
+
+
+
+
+
+
e
-
E = -V/e
W = V • Q = C • V2
C = ε • S/e
Q=C•V
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 MKSA 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 value
I=CωV
i leads v by 1/4 of a cycle
V
I
I
ϕ=- π
t
2
V
ωt
Advanced quadrature current I = C ω V
ϕ is the phase displacement of the voltage relative to the current
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Improving the power factor of an electrical installation
INDUCTANCE (SELF-INDUCTANCE) WITH A DIRECT VOLTAGE
REMINDERS (cont’d)
The passing of current through a turn creates a magnetic flux density B
in 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.
V
The voltage at the terminals of the reactor is: v = L • di/dt
I
L = coil inductance in Henry (H).
INDUCTANCE (SELF-INDUCTANCE) WITH AN ALTERNATING VOLTAGE
B
The magnetic flux density B is alternating.
i = I0 • sin ωt
v = L • di/dt = Lω I0 cos ωt
As an rms value
V=LωI
v leads i by 1/4 of the cycle
V
I
V
t
I
ϕ=+ π
2
ωt
Delayed quadrature current V = L ω I
IMPEDANCE
R
L
Z
C
z = R + jX
with
equal to
jX = jLω + (- j 1/Cω)
X = Lω - 1/Cω
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Improving the power factor of an electrical installation
REMINDERS (cont’d)
PHASE DISPLACEMENT OF THE CURRENT RELATIVE TO THE VOLTAGE
U
jXI
ϕ
RI
I
u=Z•i
ϕ phase displacement of the current relative to the voltage
tgϕ = X/R
u(t) = Uo sin ωt
i(t) = Uo/IZIsin (ωt - j)
APPARENT POWER
This is the product of the voltage and the current.
For single phase
For three-phase
S
S = UI
—
S = √ 3 UI
expressed in kVA
Q
ϕ
P
ACTIVE POWER
This is the product of the voltage, the current and the cos ϕ
for single phase
for three-phase
P = UI cos ϕ
—
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
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Improving the power factor of an electrical installation
REMINDERS (cont’d)
REACTIVE POWER
This is the product of the voltage, the current and the sin ϕ
for single phase
for three-phase
Q = UI sin ϕ
—
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(kVAR)/P(kW)
For MV 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
Generation
G
Consumers
Reactive energy
R
Public and private
distribution networks
Active energy
Losses
Mainly joule
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!
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c
c
c
increase in voltage drops, etc.
increase in Joule losses (the current transmitted is even higher)
and above all financial losses according to the billing used in the country.
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Improving the power factor of an electrical installation
COMPENSATION
WHY COMPENSATE?
In designing our electrical installation, we can note the following energy
exchanges:
Active energy
Reactive energy
Production
Losses
Consumption
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).
Use 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.
Active energy
Reactive energy
Production
Losses
A local reactive energy source
avoids transmission of this energy
Consumption
Adding a reactor to the capacitors enables filters to be created and limits
harmonic transmission.
Capacitor use near the generators enables optimization of their operation.
Active energy
Reactive energy
Production
Losses
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Optimization of energy
production management
Consumption
page 6
Improving the power factor of an electrical installation
The additional consequences of compensation are therefore to:
COMPENSATION (cont’d)
enable more economical sizing of the installations (lines, wires, transformers,
circuit breakers);
c
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 COMPENSATION
The following graph corresponds to a series compensation arrangement.
The analog graph exists in shunt compensation but in terms of current.
j.Lω.I
-
j.I
Cω
U
ϕ
R.I
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.
j.Lω.I
-
j.I
Cω
j.Lω.I
-
j.I
Cω
j.Lω.I
-
j.I
Cω
U
ϕ=0
ϕ
R.I
U
R.I
Partial compensation
GOOD
Total compensation
theorically IDEAL
The reduction of cos ϕ is adjusted
as a function of the load by using
multi-step capacitors.
Difficult to implement especially
if the load varies.
ϕ
R.I
U
Over-compensation
DANGER
Dangerous overvoltages
the equipment.
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Improving the power factor of an electrical installation
PUTTING IT IN PRACTICE
THE SERIES CAPACITOR
Series compensation enables the addition of a capacitive voltage drop
which 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.
Z ligne
C
U1
U2
Charge
ϕ
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).
ed
ensat
p
m
o
c
∆U un
ed
ensat
p
m
o
∆U c
jX.l
ted
pensa
m
o
c
n
sated
U1 u
U compen
-
1
ϕ
U2
I
j.I
Cω
R.I
For a given U2 voltage, compensated U1 is lower than uncompensated U1.
For a given U1 voltage, compensated U2 is higher than uncompensated U2.
At the end of the line, the total voltage drop is lower.
At full load, the current is inductive.
With no load, the line is capacitive; it is necessary to regulate compensation
to avoid overcompensation.
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Improving the power factor of an electrical installation
PUTTING IT IN PRACTICE
THE SHUNT CAPACITOR
(cont’d)
For utilities 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.
Z line
C
U2
Load
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.
Z line
Optimizes
purchasing
costs
Limits joule losses
and optimizes
purchasing costs
C
Load
Adding a shunt capacitor bank enables the current flowing in the lines
to be reduced.
ϕ
U2
= line current
Z before
ϕ corrected
I
U2
j Cω U2
compensation
The current flowing in the line is lower, the Joule losses are therefore reduced.
Depending on how reactive energy is billed by utility companies, energy
costs can be greatly reduced.
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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 threshold
value as in Spain or in Egypt;
c a kVA contract with a threshold of overbilling over a period of time as
in the UK and in Canada.
page 9
Improving the power factor of an electrical installation
PUTTING IT IN PRACTICE
For production plant
(cont’d)
We can consider three situations:
The main production plant permanently supplies a network
In this case, it 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.
Alternators are guaranteed for operating conditions at a given cos ϕ 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 network
This 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 generator
The 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.
INSTALLING SHUNT
CAPACITORS
The following diagram summarizes the various installation possibilities.
➊
➋
➌
➌
➍
➏
LV customer
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➎
MV customer
➏
MV customer
➊ HV bank on an HV distribution network
➋ MV bank on an MV distribution network
➌ MV bank for an MV customer
➍ LV bank set or regulated for an LV customer
➎ LV bank for an MV customer
➏ LV bank for individual compensation
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Improving the power factor of an electrical installation
APPLICATION EXAMPLE
For an installation with two motors:
800 kVA
400 V 50 Hz 3 phase
2 motors 250 kW
cos ϕ = 0.75
The power necessary to supply the two motors is:
S = P/(η x cos ϕ) or η = 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 cos ϕ1 = 0.75
tg ϕ1 = 0.88
tg ϕ2 = 0.43
if cos ϕ2 = 0.92
therefore (tg ϕ1 - tg ϕ2) = 0.88 - 0.43 = 0.45 = (Q1 Q2)/P = Qc/P
Qc is the power necessary for the capacitor bank.
Qc = 250 x 2 x 0.45
= 225 kVAR
Qc
S1
In this theoretical example,
we cannot overload the transformer
and therefore can economize the
billing of 225 kVAR reactive energy.
S2
Q
ϕ2
ϕ1
P
Generally passing from cos ϕ = 0.8 to 0.93:
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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.
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Improving the power factor of an electrical installation
HOWEVER
COMPENSATING CAN BE
A SOURCE OF TROUBLE
THE MOTOR WHICH BECOMES A GENERATOR
To supply the reactive energy that is needed, a capacitor bank has been
installed 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.
Qc
This self-excitation can cause overvoltages greater than the maximum voltage
levels for the network.
M
To avoid this problem, it is necessary to verify that the bank power is less than
the power required for excitation.
Pn
Io
Ic capacitor current = 0.9 Io no-load motor current
(see the technical data sheet Rectiphase 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
M
Io
Qc
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 compared
with the compensating action of capacitors (see cahier technique CT142).
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Improving the power factor of an electrical installation
HOWEVER
COMPENSATING CAN BE
A SOURCE OF TROUBLE
TOO MUCH OF A GOOD THING: OVERCOMPENSATING
I
π
ϕ =- 2
(cont’d)
jX.l
U1
R.I
U2
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
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, MUST FLOAT
The potential of the neutral point must remain floating to avoid an
overvoltage in one of the steps of the capacitor bank in case of line failure
or ferro-resonance. Generally, capacitor banks are wired:
Double star if:
network U > 11 kV and Qc > 600 kVAR
network U < 11 kV and Qc < 1200 kVAR
Delta if:
network U ≤ 11 kV and Qc ≤ 1200 kVAR
In MV, in the double star arrangement, the circulating of current between
two neutral points is constantly monitored to detect internal capacitor faults.
Use of capacitor banks enables an installation to be sized economically.
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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.
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Improving the power factor of an electrical installation
DEALING WITH
HARMONIC POLLUTION
Electrical power is generally distributed in the form of three voltages making
up a three-phase sinusoidal system. One of the parameters of this system is
the wave form which must be as near as possible to sinusoidal.
(see cahier technique 152)
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 effects
Vibrations, 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.
Resonance 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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Improving the power factor of an electrical installation
DEALING WITH
HARMONIC POLLUTION
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.
(cont’d)
(see cahier technique 152)
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.
HARMONIC FILTERS
network
Harmonic
generator
Other current
consumers
F5
F7
F11
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.
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Improving the power factor of an electrical installation
DEALING WITH
HARMONIC POLLUTION
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
(cont’d)
(see cahier technique 152)
Distortion
Ih/I1%
40%
Without filtering
With filtering
30%
20%
10%
0%
1
3
5
7
11
13
17 19
Harmonic number
63 kV
Scc = 200 MVA
1
Sn = 50 MVA
Scc = 330 MVA
Filters
11 kV
1.1
Gh = 40 MVA
P
20 MW
cos ϕ 1 = 0.75
cos ϕ 2 > 0.9
Q to calculate
CHOICE OF BREAKING
DEVICES
F5
F7
F11
5.487 4.403 3.66
Total Q =13.55 MVAR
Control of capacitor banks requires the use of a breaking device in which the
nominal current is derated in order to avoid heating by harmonics.
Generally, derating is to the order of 30%.
(see MT partenaire, chapter B-3-2).
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