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The Story of
Electrical Energy, Pt.14
This month, we look into the methods that make
electrical energy use more efficient. For optimum
voltage regulation and lowest running costs, the
current and voltage need to he in phase or close
to it. If they are not in phase, the energy losses
are higher than they should he.
By BRYAN MAHER
Inductive electric machines and
appliances draw lagging outofphase
currents . Naturally, we would like the
power station alternators to operate
with minimum electrical losses and
best power efficiency. And the same
may be said for the high voltage transmission lines and transformers. To
this end, the loads supplied should
ideally draw a current in phase with
the voltage. Why should that follow?
Just read on.
In most industries, AC squirrel cage
induction motors form a major percentage of the load. Induction motors
produce mechanical rotation by the
interaction of the stator's rotating magnetic field with the conductors of the
rotor. Being highly inductive , such
motors draw currents which rise later
than the voltage in each cycle. This is
illustrated in Fig.1. In this diagram,
the current is lagging the voltage by
some phase angle <j>; ie, the current
peaks come after the voltage peaks
and so on.
Resistive loads such as heaters, ovens, filament lamps and electrolytic
baths (used for plating, etching, etc)
draw a current which is exactly in
phase with the voltage as shown in
Fig.2(a). In other words, the current
and voltage rise and fall together
throughout each cycle. (The word
"phase" originally meant "a division
of time".)
Inductive loads lag. Because the
back voltage (or backEMF) generated
by an inductance opposes the applied
voltage, a pure inductor takes a cur
SUPPLY VOLTAGE
 . ._ ✓ MOTOR CURRENT
''
'
TIME AND
ANGLE
i LAG ANGLE
ONE CYCLE =360 =20 MILLISECONDS (50Hz)
Fig.I: this diagram illustrates the fundamental nature of an
induction motor load; its current waveform lags the voltage
waveform. This causes problems because the distribution
system must supply additional current.
SILICON CIIIP
0
Power & power factor
The power as_sociated with a current is equal to the product of voltage,
current and the cosine of the phase
angle. This is expressed by the following formula:
P = V.I.cos<j>
The cos<j> term is also called the
power factor (PF). Readers who remember their trigonometry may recall that the cosine of zero degrees is
unity and that cos goo = o.
Since there is zero phase angle between the voltage and current through
resistive loads, their power factor is
unity (because cos O = 1). This means
that the power dissipated in a resistive load is simply the product of the
voltage multiplied by the current.
By contrast, all purely reactive (inductive or capacitive) loads do not
use any power at all! This is because
the cosine of their go phase angle
(lag or lead) is zero and therefore they
have zero power factor.
0
<I>
86
rent which is delayed by go with
respect to the voltage, as shown in
Fig.2(b).
By contrast, capacitive loads lead.
Because capacitive current is proportional to the rateofchange of voltage,
a pure capacitor draws a current which
is exactly goo ahead of the voltage.
Fig.2(c) shows that capacitor current
is greatest where the voltage is changing at maximum rate; ie, at the zerocrossing crossing points on the voltage curve.
Real machines
In the real world, all inductive components have some resistance in addition to their inductance. Thus, iron
cored coils have a power factor greater
than zero but less than unity. Motor
windings show both simple ohmic
resistance and also inductance. But
there's more here than meets the eye.
A motor's job in life is to convert
electrical current into mechanical ro

/
CURRENT IN PHASE
.,,. 
' /
"
CURRENT LAGGING
'\
/
/
I
'\
I
90' LAG
'
'
j",_
90' LEAD
CURRENT LEADING
Fig.2: this diagram shows the phase difference between the voltage
and current for three types ofload: (a} for resistive loads, the voltage
and current are in phase; (b} for pure inductive loads, the current
lags the voltage by exactly 90°; and (c} for pure capacitive loads, the
current leads the voltage by exactly 90°.
tation of the shaft. The total induction
motor current can be visualised as
having five components:
(1). An inductive component to provide the magnetic field. This component has 90° lag angle;, ie PF = 0.
(2). The simple ohmic resistance of
the copper windings.
(3). The eddy current iron losses.
(4). · The mechanical friction and
windage losses .
(5). The mechanical power output
at the shaft.
Note that the last four components
all have a power factor of 1.
The first four of the above components are reasonably constant. However, that fifth component of motor
current clearly varies with any
changes in mechanical load.
Factory example
Suppose a factory receives a 3phase
supply at 11kV. This company would
provide its own 11kV/415V transformer for all its machines. Perhaps
this installation contains 100 induction motors.
Let's further suppose that the average motor current is 134.54A. If we
made careful measurements, we might
find the power or inphase component to be 100A, while the magnetising or lagging outofphase current
component is 90A.
Assuming this motor to be typical
of all 100 machines, the total factory
current supplied by the transformer
is 13,454 amps, which amounts to
quite a sizeable installation. The vector diagram of Fig.3 shows how the
current components relate.
In this example, the 42° phase angle is typical for the whole factory,
where some motors are on full load
and some are lightly loaded. If a motor is on no load, the inductive component of current predominates, so
the motor current would have a large
lag angle (about 80°} and a low power
factor (around 0.2).
But on full load, the power component (number 5 in the list above) exceeds all others. Thus, the total motor
current would have a small lag angle
(around 30°). The full load power factor would be about 0.85.
You might wonder what lag angle
and power factor has got to do with
the price of energy. In fact, the factory
electricity bill is charged only on the
inphase or power component. The
meter measures kilowatthours.
The kilowatt is truly a measure of
power and the kWh reading is the
energy used at a certain power over
some number of hours. None of these
indicate anything of the reactive outofphase current component, because
its power factor equals zero.
So why should the consumer worry?
Why should the factory manager care?
He should and does. For consumers
like our industrial example, the outofphase component of current is a
very big part of the total amperes flowing into and through the factory's
11kV/415 transformer.
We call the product of voltage and
outofphase component of current by
the name Volt Amps Reactive, or VAR.
Naturally a thousand of these we call
kVAR, (kiloVARs},oramillionMVAR,
(megaVARs).
As we have already seen, the main
factory transformer must supply 415 V,
3phase, at 13,454A. That's quite a big
unit and its size is expressed by the
product of voltage and line current.
This product is (✓ 3 x 415V x 13,454A)
= 9.6707 million volt amps. This
would normally be expressed as
9.67MVA.
Looking back, we see that MW is
simply the product of MVA and PF.
However, if built to provide almost
10MVA, that factory transformer is
likely to be a lot more expensive than
"""'go""'oo""A""M""Ps'"'1""'Lo""u=TO:::F""P""HA""SE:,,8
This is a vector diagram
showing an induction motor
load in which the phase angle
bet\\!;een the load current and
the applied voltage is 42 °. If a
suitable bank of capacitors
was connected in parallel, the
phase angle between current
and voltage could be reduced
to zero and the resulting
current drawn from the
distribution system also
reduced.
OCT0BER1991
87
would have to flow in the transformer's 1 lkV primary, in the state grid
system and all the way back to the
alternator stator windings in the power
station. Without power factor correction somewhere in the system, all
power line conductors, circuit breakers, transformers and metering equipment would have to be of higher current capacity, just to supply that outofphase current component for which
the customer would otherwise pay no
extra.
Therefore, most supply authorities
apply penalty charges to the electricity bill to cover the cost of the additional plant they must provide to cover
low power factor loads. Usually a
"maximum demand" recording ammeter is installed and the price per
kWh of power used is increased as
a penalty for currents above a predetermined value.
Control of capacitors
You might wonder what happens if
half the motors are switched off for
some of the time? In this circumstance,
This is a large outdoor bank of capacitors for power factor correction. There are
three vertical banks, one for each phase of the mains supply.
it needs to be. Observe that only the
10,000A inphase component of the
current produces any power. The other
component, the 9000A magnetising
current (although vitally necessary to
form the magnetic fields) produces
no power. Neither does it consume
any power from the supply.
Wouldn't it be nice if the transformer only had to supply the 10,000A
inphase component of current? This
can be done by adding capacitors to
the circuit.
Capacitors
Recall that capacitors take a leading current; ie, a perfect capacitor takes
a current which leads the voltage by
90°. Fig.2(c) illustrates this. Ifwe add
enough capacitors to the secondary
windings of the transformer, we can
88
SILICON CHIP
cancel out the lagging current of the
motors.
Because it is a 3phase system, we
will require three identical capacitor
banks. The photos show some typical
installations.
So by adding suitable capacitors to
the installation, we have reduced the
transformer secondary current from
13,454 amps down to 10,000 amps.
This enables a transformer rated at
around 7MVA to be used instead of
one rated at lOMVA. That's a big reduction in size and cost! Running costs
are also reduced. Smaller transformer
currents mean smaller losses and
therefore less continual costs.
Penalty charges
There are other benefits too. Without capacitors installed, extra current
Banks of power factor correction
capacitors can be very large or
relatively modest as with these units
made at Asea Brown Boveri's
Capacitor Division at Lilydale in
Victoria.
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You can now purchase the
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Here's what you get:
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The induction motor is the cause of most power factor problems in the AC
power distribution network, along with fluorescent lamps. Because it is an
inductive load, the motor current lags the AC voltage waveform and that means
extra current has to be provided.
the remaining motors only take about
4500A of outofphase lagging current
but the capacitor bank would still be
taking its full 9000A of leading current! This would amount to having
overcompensation and so the total
factory current would be higher tha.n
necessary, with a leading power factor!
There are three ways to prevent such
a situation:
(1) Rather than have one large bank
of capacitors for the whole factory, we
could use separate capacitors for each
motor. This way, when the motor is
switched off, so too is the capacitor.
This method can have a severe disadvantage, though. Each time the
motor circuit breaker is opened, if the
capacitor and motor inductance find
resonance at the wavefront frequency,
a large overvoltage spike may be generated. In extreme cases, this has been
known to puncture the motor winding insulation. When next the motor
is switched on, explosive breakdown
can occur.
(2) Alternatively, the large bank of
capacitors may be divided into sections, each brought on line as required
according to the number of motors in
use. Computers are now used to monitor the power factor and control the
switching of capacitor banks.
(3) Finally, another method of power
factor correction involves using a unloaded synchronous motor. However,
that m ethod is beyond the scope of
this article.
Fluorescent lights
It is not only induction motors that
produce lagging power factor problems. Fluorescent lamps also have a
lagging power factor, because of the
current limiting inductor (known as a
ballast). This typically results in a
power factor of between 0. 7 and 0.8.
When you consider the thousands of
fluorescent lamp fittings used in all
commercial buildings, factories and
public buildings such as schools and
hospitals, the leading power factor
becomes a major problem for the supply authorities.
For this reason, in commercial and
public buildings, the supply authority requires that each fluorescent lamp
fitting contain a power factor correction capacitor. This is connected in
parallel with the 240VAC input. Power
correction capacitors are not required
for fluorescent lamps used in homes,
however.
Acknowledgements
Grateful thanks to the NSW Electricity Commission and to ASEABrown Boveri for data and photos and
permission to publish.
SC
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OCT0BER1991
89
