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.
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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
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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
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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
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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
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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.
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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
Active energyReactive energyProduction
Optimization of energyproduction management
Losses
Active energyReactive energyProduction
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.
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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.
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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.
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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.
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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
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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.
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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
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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.
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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.
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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).