Download Power Factor Improvement Experiment # 4 Power Factor

Salman bin AbdulazizUniversity
College of Engineering
Electrical Engineering Department
EE 2050Electrical Circuit Laboratory
Experiment # 4
Power Factor Improvement
Objectives:
1. To introduce the concept of power factor and method of power factor improvement for a single phase
inductive loads.
Equipments:
1. One Fluorescent lamp. (110 V, 20 W)
2. Two multi-meters.
3. One Wattmeter.
4. One decade capacitor box.
Note:
Please record any damage, if exists any while performing the experiment. You can also write the
th
difficulties you are confronted with when using the equipment, and your suggestions and critics
related with the equipment you used.
Background:
1. Power in resistive and reactive AC circuits
Consider a circuit for a single
single-phase AC system shown in Fig. 1, where a 120 volt, 60 Hz AC
voltage source is delivering power to a resistive load.
Fig. 1. Ac source drives a purely resistive load.
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In this example, the RMS current to the load would be 2 A. Therefore power dissipated at the
load (active power) would be (2)2×60 = 240 W. Because this load is purely resistive, the current
is in phase with the voltage. Fig. 2, shows the voltage, current and power waveforms.
Fig. 2. Current is in phase with voltage in a resistive circuit.
Note that the power waveform is always positive, this means that power is always being
dissipated by the resistive load, and never returned to the source.
For comparison, let’s consider a simple AC circuit with a purely reactive load as shown in Fig. 3.
Fig. 3. AC circuit with a purely reactive (inductive) load.
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Fig. 4 Current lags voltage with 90º in a pure inductive circuit.
As seen from Fig. 4, power (p) alternates equally between cycles of positive and negative. This
means that power is being alternately absorbed from and returned to the source.
Now, let’s consider an AC circuit with a load consisting of both inductance and resistance as
shown in Fig. 5.
Fig. 5. AC circuit with both reactance and resistance.
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Fig. 6. Current, voltage and power waveforms
As with any reactive circuit, the power alternates instantaneously between positive and negative
values over time. In a purely reactive circuit, alternation between positive and negative power is
equally divided, resulting in a net power dissipation of zero. However, in circuits with mixed
resistance and reactance like this one, the power waveform will still alternate between positive
and negative, but the amount of positive power will exceed the amount of negative power. In
other words, the combined inductive/resistive load will consume more power than it returns back
to the source.
2. True, Reactive, and Apparent power
We know that reactive loads such as inductors and capacitors dissipate zero power, yet the fact
that they drop voltage and draw current gives the deceptive impression that they actually do
dissipate power. This “ghost power” is called reactive power, and it is measured in a unit called
Volt-Amps-Reactive (VAR), rather than watts. The mathematical symbol for reactive power is
the capital letter Q. The actual amount of power being used, or dissipated, in a circuit is called
true (active) power, and it is measured in watts (symbolized by the capital letter P). The
combination of reactive power and true power is called apparent power, and it is the product of a
circuit’s voltage and current, without reference to phase angle. Apparent power is measured in
the unit of Volt-Amps (VA) and is symbolized by the capital letter S. There are several power
equations relating the three types of power to resistance, reactance, and impedance:
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The three types of power can be represented by a triangle shown in Fig. 7.
Fig. 7. Power triangle relating apparent power to true power and reactive power.
From Fig. 7, the angle of this “power triangle” graphically indicates the ratio between the
amount of dissipated (or consumed) power and the amount of absorbed/returned power. It also
happens to be the same angle as that of the circuit’s impedance in polar form. When expressed as
a fraction, this ratio between true power and apparent power is called the power factor for this
circuit.
It should be noted that power factor, like all ratio measurements, is a unitless quantity.
For the purely resistive circuit, the power factor is unity (perfect), because the reactive power
equals zero. Here, the power triangle would look like a horizontal line, because the opposite
(reactive power) side would have zero length. For the purely inductive circuit, the power factor is
zero, because true power equals zero. Here, the power triangle would look like a vertical line,
because the adjacent (true power) side would have zero length.
3. Improving the System Power Factor
Consider an RL circuit whose circuit impedance can be expressed as ZL = R+j XL = ZL ∠θL.
Suppose the expression for the supply voltage is v = √2 V sin ω t , the expression of current will
be √2 I sin (ω t −θL), where I = V / ZL, is the rms value of current and θL is its phase angle. The
instantaneous power P = vi = 2 VI sin (ωt) sin (ω t –θL) can be expressed as:
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From the above equation, it is clear that the first term is constant while the second term is a
cosine wave which has a frequency of 2ω and a peak value of VI. The average value of this term
is zero. Thus the average value of power delivered to load, Pav , is
The term cos θL in the above equation is called power factor (PF) of load. If the load PF is less
than unity it may be either capacitive (i.e. leading) PF or inductive (i.e. lagging) PF.
However, in practice almost all the loads are inductive with lagging PF. When the PF of a load is
less than unity, it takes more current for the same power from the supply. This large current
requirement will cause more I 2R losses and IR drop in the supply line. Moreover, large current
will demand greater current capacity for the generator and associated systems. Therefore, it is
desired to correct the PF to unity. Lagging PF can be improved by connecting capacitors across
the inductive load as shown in Fig.8. The shunt capacitor will not absorb real power but instead
it will take leading current from the line to neutralize the lagging current component of the load,
thereby making the net supply current less and hence improving the system PF.
Fig. 8. Inductive load with a compensating capacitor
For the circuit of Fig.8, if the switch S is off, load current IL is given as
Where ZL2 = R2 + XL2.
The first component of the current IL given above is called active component while the second
component is called reactive component of current. This reactive component which is lagging the
supply voltage by 90° can be neutralized if a capacitor is connected across the load, in case of the
switch S is turned on, which draws a reactive current of similar magnitude but leading the supply
voltage by 90° (i.e. IC = j V XL / ZL2).
Thus, with the capacitor connected parallel to the load, the supply current IS is given as
Since IC = j V /XC = j V (2 π f C). Thus, to get a PF of unity, we must put C such that
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If C is less or more than that given by the above equation, the resulting power factor will be less
than unity and will be lagging or leading respectively.
Figure.9 (a) depicts the phasor diagram of voltage and current whereas Fig.9 (b) shows the
power diagram. Both of these diagrams show how the leading current or leading VA of the
capacitor neutralizes the lagging reactive current or lagging reactive volt ampere of the load,
thereby, improving the PF.
Fig. 9. Improving PF using capacitor
In both of these diagrams we see that initial or load phase angle, θL (before compensation) is
more than the final or supply phase angle, θS (after compensation), i.e.
Procedure
1. Connect the circuit as shown in Fig.10.
Fig. 10. Circuit used in procedure
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2. When the switch S being OFF, record the reading of voltmeter (V), ammeter (I) and
wattmeter before connecting the capacitor C in Table 1.
3. For the fluorescent lamp, without capacitor, estimate the load resistance R, inductive reactance
XL, impedance ZL, phase angle θL and PF. Record the value of PF in table 1. Calculate the value of
C required to make PF = 1.
C = ……………… (µF)
Table 1. Reading of voltmeter, ammeter and wattmeter at different values of C.
Capacitor C
(µF)
Not connected
Voltage (V)
Current (A)
Power (W)
Power Factor PF
1.0 (µF)
2.0 (µF)
3.0 (µF)
3.2 (µF)
3.4 (µF)
3.6 (µF)
3.8 (µF)
4.0 (µF)
4.2 (µF)
4.4 (µF)
4.6 (µF)
4.8 (µF)
5.0 (µF)
5.2 (µF)
5.4 (µF)
5.6 (µF)
5.8 (µF)
6.0 (µF)
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4. Switch ON the switch S to connect the capacitor C parallel to the load with a capacitance
value of 1.0 µF, and record the reading of voltmeter, ammeter and wattmeter again. Then
calculate the load PF then record in Table 1.
5. Increase the value of the capacitor in steps given in Table 1, and record the required readings
until you get a point where the current is minimum.
6. Plot the relationship between the load power factor PF and the capacitor value. From this curve,
Find the value of C at which the PF is maximum.
7. Compare this value with that calculated in step 3.
Load PF
C (µF)
Fig. 11. Power factor vs capacitor values
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