University of Rochester
Department of Electrical & Computer Engineering
ECE111
Laboratory #9
Power Factor Correction
week of 30 Nov 2008
--------------------------------------------------------------------------------------------------------Your write-up must provide a detailed description of the exercise, with diagrams of circuits and a
description of procedures. Your lab TA must sign and date each page of your lab notebook.
Separate from the report, include a typed abstract that meets the requirements for a good
technical abstract. Your grade will be based in part upon conciseness, grammar, and spelling.
Lat e wor k will not be acc ept ed.
--------------------------------------------------------------------------------------------------------0.
Laboratory prepa ration
Read closely sections 11-5 to 11-9 of the ECE111 text, paying particular
attention to example #11.15 found on pg. 482.
I.
Power factor of AC circuits
Consider the AC circuit shown in Fig. 1, consisting of a series L-R plus a
capacitor C connected in parallel.
I
+
V
R
C
L
Inductive
load
Figure 1.
AC circuit consisting of inductive load and shunt capacitor C.
The series L-R represents a load such as an induction motor. The power factor
(without the capacitor) is PF = cos φ (lagging), where φ = -tan-1(ωL/R). In terms
of the rms current I, the real and reactive power quantities are, respectively, PR =
RI2 and QX = +XI2. The reactive power does not contribute to the time-average
real power consumed by the load, but does increase the required current-carrying
capacity (ampacity) of the electrical wiring. The higher current flow results in
inefficiency because of ohmic losses in the wiring while the need for heavier wiring
increases the capital cost of the distribution system. Power engineers, striving to
provide electrical service to such loads at minimum cost, try to "correct" the
power factor by adding capacitance in parallel with the load. The capacitance is
chosen so that its reactive power: QB = -BV2 (where V = rms voltage and B = ωC
-2-
is capacitive susceptance) at least partially cancels out the reactive power to the
inductor: i.e., QB + QX ≈ 0. Power factor correction works because the current
flowing through the capacitor is 180° out of phase with respect to the reactive
component of the load current. As a result, the net current to the combined
network, consisting of the LR load and the parallel capacitor, can be reduced, as
long as the capacitance is correctly specified.
II.
Experimental procedure & lab questions
A.
Assemble the series R-L circuit shown in Figure 2, using RL = 1.8 kΩ, LL
= 30 to 60 mH, and Rm = 100 Ω. The resistor Rm is chosen to be small
compared to the load impedance, that is, Rm « |RL + jωLL|. As long as
this inequality is maintained, then the resistor serves as a convenient
method to measure the load current i(t) = vR(t)/Rm without significantly
disturbing the voltage across the load.
i(t)
+
shunt capacitor is
to be added in part E
R
~
vs (t)
Figure 2.
L
Rm
C
+
vR (t)
-
Laboratory setup for part II of lab.
B.
Set vs(t) = 2cos(2π•104) and then measure the peak value of i(t) and
the phase angle φ between vs(t) and i(t). What is the power factor? Is
it leading or lagging?
C.
Ignoring the effect of the resistor Rm on the network, calculate
theoretical predicted values for φ and the peak value of i(t) and
compare to your measurements. Explain any discrepancies observed
between the calculated and measured results.
-3-
D.
Plot the current phasor with respect to the voltage reference.
E.
Now add shunt capacitance to the network as shown. Use values for C
as follows: C ≈ 1000 pF, 2200 pF, 4700 pF, 0.01 µF, 0.022 µF, and
0.047 µF. Measure φ and the peak magnitude of the current for each
capacitance value.
F.
From your experimental data, calculate the power factor PF and tabulate
these results, along with all experimental data from E above. Use your
data to plot PF versus C. What value of C has given you the minimum
current i(t)?
G.
Now use the equation QB + QX = 0 to calculate the value of C at which
the power factor is corrected to unity, that is, PF = 1.0. To do this, you
must correctly calculate the current through the L-R load. Compare
your result to the experimental results from part F.
H.
What is the relationship between maximum power transfer and power
factor correction?
III. 60 Hz powe r factor corre ction of a choke coil
Warning: In this section of the laboratory, you will be working with potentially
lethal voltages and currents. Follow all procedures exactly and do not touch any
circuit or re-wire anything unless the Variac supply is unplugged!!
fuse
120 v
60 Hz
~
wattmeter
variac
Figure 3.
A
•
V
LOAD
•
12 µF 12 µF 12 µF
Setup for 60 Hz power factor determination.
The setup for this exercise is shown in Fig. 3. A variable autotransformer
(Variac®) provides adjustable 60 Hz AC to a load consisting of a choke (inductor
with series resistance). Several shunt-connected capacitors may be switched into
the network to adjust PF. This network is instrumented with an AC voltmeter, AC
ammeter (both reading rms), and a power meter (reading average real power).
With this arrangement, the power factor can be determined directly using the
equation: PF = P/VrmsIrms.
-4-
A.
First, characterize the choke by measuring its power factor PF and then
calculating Qload with the voltage set to 80 v-rms. Is power factor
leading or lagging? Explain how you arrived at your answer.
B.
Use the results from part III.A to determine the resistance R and
reactance X for the load. Specify units for these quantities.
* TO AVOID OVERHEATING CHOKE COIL, DO NOT EXCEED 80 WATTS *
C.
Adjust the Variac so that the wattmeter reads 80 watts. Record
voltage and current for all the possible values of shunt capacitance that
can be switched in: C = 0, 12 µF, 24 µF, and 36 µF.
D.
Construct a table containing all data from the above measurements plus
all the calculated results shown below. The measured and calculated
values for current, Imeas and Icalc, should agree within measurement error.
C(µF)
Vmeas
Imeas
Pmeas
PFcalc
Qcalc
Icalc
0
12
24
36
Use the following expressions to fill out the table: PFcalc = Pmeas/VmeasImeas,
Qcalc =
2
,
(VmeasI meas) 2 - P meas
I calc =
2
Qcalc @ C = 0, and Qc = - !CVmeas
.
E.
2
P meas
- (Qload +Qc) 2 Vmeas, with Qload =
Note: ω = 2π•60 ≈ 377 rad/sec.
Write a 150 word paragraph explaining the economic benefit of power
factor correction in the delivery of electrical power and presenting an
argument that it is a GREEN concept.