•
Use the provided RESULTS file for the values to use in the report
The lab report should concentrate on meaningful results, discussion, and
conclusions.
•
Must include an:
○ Abstract
Objective
○
Introduction
○
Procedure
○
о
Write a brief explanation of the procedure, (DO NOT JUST COPY THE
PROCEDURE FROM THE PROVIDED MANUAL).
Measurements and Results
Discussion of results and Analysis
Answer the highlighted questions in the lab manual
Experimental results with theoretical results and including error
analysis.
Discussion of the agreement or differences between the measured
results and the values expected from theory.
Conclusion
Meaningful summary of what you learned from the experiment./n Chapter 6 - Experiment 4
Measurement of AC Power
In this lab we will use the D52-2066 Multi-Function Digital Meter to measure the AC power delivered
to a resistive load, an RC load and an RL load. The wattmeter measures the apparent power delivered
to a load, and the power factor. It is connected so that the current in the load flows through the
wattmeter, and the voltage across the load is applied across the wattmeter. The digital meter reads the
RMS voltage across the load and the RMS current flowing through the load. The wattmeter accounts
for the phase of the current relative to the voltage to evaluate the power factor.
This lab uses high-power equipment. The Variac is an AC voltage source at a frequency of 60 Hz, and
can supply voltages up to 120 volts RMS, and currents of more than 1 amp. The high-power resistor in
the experiment can dissipate up to 150 watts. Please follow the safety instructions carefully. Always
turn the Variac down to zero to turn off the power before changing the wiring of your circuit.
6.1 Power Quantities
Consider a load consisting of a resistance R in series with an inductance L. The impedance is
Z = R+jwL= √R² + (WL)² 20,
= tan−1
where the angle of the impedance is 0
across the load, then the (phasor) current that flows is
V
(2). If we apply a (phasor) voltage V = VRMS40°
VRMS
I
=
= IRMS (-0)
Ꮓ R² + (WL)² 20
where IRMS
=
VRMS
√R²+(WL)²
The complex power is
Z
=
S = VI* = (VRMS<0)(IRMS≤(-0))* = VRMS!RMS <0 = VRMSIRMS COS 0 + jVRMS!RMS Sin 0.
The complex power is expanded as S
=
PAV+jQ, where the average power is
PAV = VRMS RMS COS 0 watts
and the reactive power is
Q
=
VRMSIRMS sin 0 vars
where "var" stands for volt-amps reactive. The apparent power is
The power factor is
Papp = Vrms¹rms Watts
pf = cos 0
where is the angle of impedance Z. With these definitions the average power is related to the
apparent power by the power factor
PAV = PAPPPf.
66
66 If we know the phasor for the voltage V and the phasor for the current I, it is convenient to calculate the
average power with PAV = Re(VI), where phasors V and I are written relative to RMS value.
Using the D52-2066 meter, we can measure the apparent power and the power factor with the circuit of
Figure 6.1. The Variac supplies the source voltage V to the wattmeter box. Connect the Variac to the
red and green banana jacks at the bottom. Connect a voltmeter across the output voltage of the Variac.
When the Variac is turned down to zero, the voltmeter reads zero volts and it is safe to change the
circuit. The load is connected between the green and the yellow banana jacks at the top. Neglecting the
voltage drop across the wattmeter the load voltage is V₁ = Vs. The load current is IL
VL The D52-
2066 meter reads the RMS load voltage VRMS, and the RMS load current IRMS. The meter reads the
apparent power Papp = Vrms¹rms delivered to the load, and the power factor pf. We can calculate the
average power as Pav = PappPf.
=
ZL
Variac
voltage
source
+
Vs
V
268.3W
0.22
228.3 500
18. 100
+51
Load Z1.
Figure 6.1 Circuit for power measurements.
6.2 Power Factor Correction
Suppose we have a factory with electric motors powering machinery. Motors tend to have an
impedance that is inductive and can be modelled as an RL series impedance
WL
where tan =
R
.
Z = R + jwL = R² + (WL)² 20
The power delivered to the RL series load is PAV = VRMS√RMS COS 0, where the
power factor is pf: = cos 0. If voltage V is fixed at 230 V RMS, then how much current is needed to
deliver an average power of PAV
factor is unity, pf = 1. The current is
=
10 kW? If the load is purely resistive with L = 0, then the power
=
PAV 10000
VRMS 230
= : 43.5 A.
IRMS
-
But if the load is resistance in series with inductance, with a power factor of pf = cos 0, then more
current is required,
IRMS
PAV
VRMSPf
If the power factor is pf = cos 66.4 = 0.401, to deliver 10 kW of power we must supply a current of
67 IRMS
=
10000
PAV
VRMSPf 230x0.401
= 108.4 A.
A much larger transformer and much thicker wires are needed to deliver such a high current from the
substation to the factory, which is expensive!
Power utilities want to keep the power factor near unity, so that the thinnest possible wires can be used.
Figure 6.3 shows a capacitor used to correct the power factor of the RL load. We can choose the
capacitor value to make the power factor of the capacitor in parallel with the RL load near to unity.
The admittance of the capacitor is Yc = jwC and of the RL load is
The admittance of Y, and Y in parallel is
YL
=
R-jwL
R²+(WL)²*
R
Y = YC + YL =
R²+(WL)²
+ j (wC -
WL
R²+(WL)²,
power factor is unity if the admittance is real. Choose the capacitor to make
The
WC
-
WL
R²+(WL)²
=
0,
which can be solved to find
C =
L
R²+(WL)²*
With this choice the power factor of the load is unity
In this experiment we will study power factor compensation for a series RL load. The frequency is 60
Hz. Consider an inductance of L = 0.2 H and a typical resistor value of R = 33 . Then the load
impedance is 82.3266.4° N, and the power factor is pf = cos 66.4° = 0.401. With a voltage of
VRMS = 120 V RMS, the current phasor is I = 1.462 - 66.4° A, and the average power delivered to
the load is PAV = Re(VI*) = Re(120(1.46z – 66.4°)*) = 70.1 watts. The student is encouraged to
verify the arithmetic! Note that in the experiment we will use a smaller voltage so that the current is 1
A RMS.
1
Y
-
To compensate for the power factor of the RL load, connect a capacitor in parallel with the load as in
Figure 6.3. To continue our example, use the formula given above to choose C = 29.52 µF. With this
value the admittance of the C in parallel with RL load is Y = 0.487 ms with zero imaginary part. The
impedance is Z = = 205 §. The impedance is real and the power factor is unity. With VRMS = 120
V, the current flowing into Z = 205 £ is I = 0.58520° A, and the power supplied to the load is PAV
Re(VI*) = Re(120x(0.5850°)*) = 70.2 watts. The current required to supply 70 watts is 0.585
amps, much less than for the uncompensated RL load, where 1.46 amps was required. By
compensating the power factor with the parallel capacitor we have reduced the current from 1.46 A to
0.585 A. Once again the student is encourage to check the arithmetic.
=
6.3 Pre-Lab Exercises
1. Read about the instruments used in AC Power measurement in Chapter 2 of this manual. The
Variac is the AC power source. Loads are constructed with an RL load unit and a power
capacitor unit. The wattmeter measures AC power and Section 2.4.4 explains how to connect
the wattmeter to the load.
68 =
2. Use LTSpice to simulate the circuit of Figure 6.3, without including the wattmeter. Use R
33 and L = 0.2 H. Use a voltage source of 120 V RMS at 60 Hz. Find the current flowing
out of the generator for the uncompensated load with C = 0 F and for the compensated load
with C = 29.52 µF. Find the power delivered to the resistor in each case. Compare the current
supplied by the generator to the uncompensated load with the current supplied to the
compensated load.
6.4 Procedure in the Lab
Because of the substantial voltages and currents used in this experiment, please follow the safety
instructions diligently.
CAUTION: THIS EXPERIMENT INVOLVES CURRENTS AND VOLTAGES THAT CAN
CAUSE ELECTRICAL SHOCK, IF IMPROPER PROCEDURES ARE USED. IN ORDER TO
AVOID SUCH HAZARDS, PLEASE OBSERVE THE FOLLOWING PRECAUTIONS:
ALWAYS return the Variac knob to its zero position after taking each set of
readings and/or before making any circuit change.
NEVER leave the R-control in the load unit in its minimum position.
DOUBLE-CHECK your circuit connections before applying power, every time.
●
•
•
Use proper patch cord lengths and colour codes.
.
DO NOT daisy-chain patch cords.
The RL load unit and the C load unit should not be moved from their positions
on top of the hutch.
With the Variac knob at its zero position, the output voltage is zero and it is safe to change the
wiring of the circuit.
6.4.1 Power Measurement
Figure 6.1 shows the circuit used for the AC power measurement experiment. The Variac is the voltage
source for the experiment. A voltmeter is connected across the terminals of the Variac to monitor the
voltage at the input to the circuit. The Variac voltage is connected to the input of the D52-2066 meter.
The load terminals are connected to the load. The load can be either a variable resistor, or the resistor in
series with an inductance, or the resistor in series with a capacitance.
Figure 6.2 shows wiring diagrams for an R load, an RL series load and an RC series load. The R load
in Figure 6.2(a) uses the RL box of Figure 2.14. The RL load of Figure 6.2(b) uses the resistor and the
inductor of the RL box in series. The RC load of Figure 6.2(c) uses the R side of the RL box and the
power capacitor load unit of Figure 2.15.
Never leave the R control knob in the zero position because this makes the load zero ohms and the fuse
in the RL box will blow when you apply the power.
69
60 R
ZL
R terminals
R-Load
R
ZL
RL -Load
(a) Resistive load
R control
knob
L terminals
R terminals
R
ZL
(b) Series RL load.
R terminals
RC-Load
R control
knob
(c) Series RC load.
Figure 6.2 Loads using the RL box and the power capacitor load unit.
Ten different RL boxes are available in the lab, with inductances of 188, 195, 197, 199, 205, 210, 210, 214,
220 and 230 mH, which are all close to 200 mH. The capacitance values indicated on the capacitor box are
accurate.
Before you start the experiment check to see that the fuses on the RL box is good.
Rheostat Calibration
Calibrate the rheostat in the RL box by measuring the resistance at each R dial setting with the
ohmmeter. The resistor has an overall resistance of roughly 100 N and we expect the resistance to
increase linearly with dial setting, from 0 to 100 2.
70
70
