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Power Circuit &Electromagnetic (EET 221/3 - Software Version)
Laboratory Module
EXPERIMENT 1
TITLE: POWER IN AC CIRCUITS
OBJECTIVES
1) To determine the active, reactive and apparent power supplied to an
inductive load by using measurements of circuit currents and voltages.
2) To improve the power factor.
EQUIPMENTS
EMS Mobile Workstation Model 8110, Resistive Load Model 8311, Inductive
Load Model 8321, Capacitive Load Model 8331, Power Supply Model 8821 and
Data Acquisition Interface Model 9062.
INTRODUCTION
The apparent power supplied to a load equals the product of the load voltage and
current. In ac circuits, apparent power is always greater than active power when
the load contains reactance, because reactive power must be supplied by the
source. The reactive power may be inductive or capacitive, but in most
electromechanical devices it will be inductive because of the inductance of the
coil in transformers and motors.
The formula for determining reactive power in an ac circuit is
Q = √ S2 – P2
Where Q = reactive power in vars,
S = apparent power in VA,
P = active power in W.
When the phase angle Φ between the voltage and current is known, active power
can be determined from the following formula:
P = EI cos Φ = S cos Φ
In ac circuits where the voltage and current are sine waves, the terms cos Φ is
called the power factor and it equals the ratio of active power to apparent power,
P/S. The actual value of the power factor depends on how much the current and
voltage are out-of-phase. If E and I are in phase (purely resistive circuit),
meaning phase angle Φ is O°, then cos Φ = 1 and the active power equals the
product EI (apparent power). When the phase angle between current and voltage
is 90° (purely reactive circuit), cos Φ = 0, therefore the active power is zero.
When a circuit contains both resistance and reactance, the phase angle Φ lies
between 0° and ± 90°, depending on whether circuit reactance is inductive or
capacitive, cos Φ has a value between 0 and 1, and the active power equals a
fraction of the apparent power.
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Power Circuit &Electromagnetic (EET 221/3 - Software Version)
Laboratory Module
Power distribution analysis of ac circuits can be simplified using the power
triangle technique and Figure 1-1 shows how P, Q and S are related. The angle
between the active power axis (x-axis) and the hypotenuse of the triangle
corresponds to the phase angle Φ. Inductive reactive power is drawn in a +y
direction and capacitive reactive power is drawn in a –y direction.
Certain texts use a convention for inductive and capacitive reactive power
opposite to the one used here. Inductive power is shown as a negative vector
quantity because inductive current lags the voltage across the inductor, and
capacitive power is drawn as a positive vector quantity since capacitive current
leads the voltage.
Figure 1-1 The Power Triangle
AC motors draw inductive reactive power from the ac power supply to create the
magnetic field which they require. In addition, ac motors absorb active power,
most of which is converted to mechanical power and dissipate the remainder as
heat. The reactive power travels back and forth between the motor and the ac
power supply and it does no useful work other than creating the magnetic field for
the motor. If a capacitor is placed in parallel with the motor and its value adjusted
so that the capacitive reactive power exactly equals inductive reactive power, the
negative reactive power of the capacitor will cancel out the positive reactive
power of the motor. In fact the reactive power will travel back and forth between
the motor and the capacitor instead of traveling back and forth between the
motor and the ac power supply. The ac power supply will no longer have to
supply reactive power, which will result in a large reduction in the current the
motor draws from the power supply. Adding capacitive reactance in this way to
lower the current (the current drawn from an ac power source) is called power
factor correction and leads to improved line regulation. Also, this allows smaller
diameter wire to be used for the transmission lines. The power factor of an ac
motor by itself is usually quite low, often below 0.7, but once the capacitor/motor
combination is in place, the power factor is substantially improved. With the
proper choice of capacitance, the power factor will be close to unity.
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Power Circuit &Electromagnetic (EET 221/3 - Software Version)
Laboratory Module
PROCEDURE
CAUTION
High voltages are present in this laboratory exercise! Do not make or modify any
banana jack connections with the power on unless otherwise specified!
1. Install the Power Supply, Data Acquisition Interface, Resistive Load, Inductive
Load and Capacitive Load modules in the EMS Workstation.
2. Make sure that the main switch of the Power Supply is set to the O (OFF)
position, and the voltage control knob is turned fully counter clockwise. Set the
voltmeter select switch to the 4-N position.
3. Set up the circuit shown in Figure 1-2. The RL section of the circuit simulates the
load of a single-phase ac motor. Connect all three sections of the Resistive and
Inductive Load modules in parallel and set R and XL to the values given. Connect
I1 and E1 as shown to measure the circuit current and voltage.
4. Ensure that the DAI LOW POWER INPUT is connected to the main Power
Supply.
5. Display the Metering application.
6. Turn on the main Power Supply and set the 24V-AC power switch to the I (ON)
position. Adjust the voltage control to obtain the value of ES given in Figure 1-2.
7. Measure the load voltage and current and the active power consumed by the
circuit. Note the results, and then turn off the power.
Figure 1-2 RL Load to Simulate an AC Motor.
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Laboratory Module
8. Use the measured values of E and I to determine the apparent power supplied to
the load.
9. Determine the power factor cos Φ, and the reactive power Q.
10. Do the values calculated in step 9 demonstrate a low power factor and a notable
amount of reactive power for the simulated motor load? Why?
11. Modify the RL circuit by adding capacitive reactance in parallel with the load as
shown in Figure 1-3. Ensure that all sections of the Capacitive Load module are
connected in parallel and that all the switches on the module are open.
12. Turn on the power and add capacitance to the circuit by closing the first switch in
each section one after the other, then the middle switches and finally the third
switch in each section, until all switches have been closed. At each new value of
capacitance, click the Record Data button to record the line current measurement
in the Data Table.
Figure 1-3 Power factor Correction by Adding Capacitive Reactance.
13. After all data values have been recorded, display the Graph screen, select I1 as
the Y-axis parameter and click the Line Graph button to observe the line current
variation curve. Does the line current increase, decrease or stay the same as
more and more capacitance is added to the circuit?
14. Is there a point at which the line current stops decreasing, and then starts to
increase again when more capacitance is added?
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Power Circuit &Electromagnetic (EET 221/3 - Software Version)
Laboratory Module
15. Carefully adjust the switches on the Capacitive Load module to obtain minimum
line current, while readjusting the voltage control as necessary to maintain the
exact value of ES given in the table. Use the settings on the module to determine
the value of capacitive reactance that produces minimum line current.
Note: You will have noticed that the line current is minimum when the capacitive
reactance equals the inductive reactance. The negative reactive power then
cancels the positive reactive power and line current is minimized.
16. With Xc adjusted for minimum line current, record the value of E, Imin and the
active power on PQS1.
17. Determine the apparent power S.
18. Calculate the power factor cos Φ, and the reactive power Q.
19. Has the reactive power consumed by the circuit decreased between step 9 and
step 18? Why?
20. Has the line current been reduced by a significant amount with the addition of
capacitance?
21. Is the active power consumed by the RL load approximately the same with and
without capacitance?Why?
22. Ensure that the Power Supply is turned off, the voltage control is fully counter
clockwise and remove all leads and cables.
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Power Circuit &Electromagnetic (EET 221/3 - Software Version)
Laboratory Module
Name: _________________________ Matrix No.: _____________ Date: __________
RESULTS & CALCULATION
Please circle the correct answer for question no. 14 & 20.
7.
E = __________ V,
I = ___________ A,
P = ___________ W
8.
S = E x I = ________________________ = ______________ VA
9.
cos Φ = P = __________ = ___________
S
Q = √ S2 – P2 = √
10.
= _____________ vars
________________________________________________________________
________________________________________________________________
________________________________________________________________
12.
Sample Numbers
1
2
3
4
5
6
7
8
9
10
Line Current, I1
A
Table 1-1 Line Current
Instructor Approval: _____________________________________ Date: __________
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Power Circuit &Electromagnetic (EET 221/3 - Software Version)
Laboratory Module
Name: _________________________ Matrix No.: _____________ Date: __________
13.
Line Current versus Power Factor
2
Line Current (A)
1.5
1
0.5
0
2
4
6
8
10
Sample Numbers
Figure 1-4 Line Current versus Power Factor
________________________________________________________________
________________________________________________________________
________________________________________________________________
14.
Yes/No
15.
Xc =
1
=
_____________________
= ______________Ω
2πfC
16.
E = __________ V,
IMIN = ___________ A,
P1 = ___________ W
Instructor Approval: _____________________________________ Date: __________
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Power Circuit &Electromagnetic (EET 221/3 - Software Version)
Laboratory Module
Name: _________________________ Matrix No.: _____________ Date: __________
17.
S = E x IMIN = _______________________ = ___________ VA
18.
cos Φ = P1 = ___________ = ___________
S
Q = √ S2 – P12 = √
19.
= _____________ vars
_______________________________________________________________
________________________________________________________________
20.
Yes/No
21.
_______________________________________________________________
_______________________________________________________________
________________________________________________________________
Instructor Approval: _____________________________________ Date: __________
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Power Circuit &Electromagnetic (EET 221/3 - Software Version)
Laboratory Module
Name: _________________________ Matrix No.: _____________ Date: __________
DISCUSSION
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
CONCLUSION
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
________________________________________________________________________
Instructor Approval: _____________________________________ Date: __________
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Power Circuit &Electromagnetic (EET 221/3 - Software Version)
Laboratory Module
Name: _________________________ Matrix No.: _____________ Date: __________
PROBLEMS
Please circle the correct answers
1. An electromagnet draws 3 kW of active power and 4 kvars of inductive reactive
power. What is the apparent power?
a.
b.
c.
d.
500 VA.
5 kVA.
50 kVA.
7 kVA.
2. What is the power factor cos Φ for the electromagnet in Question 1?
a.
b.
c.
d.
0.75.
1.33.
0.60.
1.00.
3. A capacitor drawing 4 kvars of reactive power is placed in parallel with the
electromagnet in Question 1. How does this effect the apparent power S and the
power factor cos Φ?
a.
b.
c.
d.
Apparent power now equals the active power and cos Φ equals 1.
Apparent power is doubled and cos Φ remains the same.
Apparent power remains the same and cos Φ decreases.
Both apparent power and cos Φ increase.
4. What is the formula used to determine reactive power Q?
a.
b.
c.
d.
Q=S–P
Q = S cos Φ
Q = EI cos Φ
Q = √ S2 – P2
5. A capacitor drawing 8 kvars is placed in parallel with an electromagnet that
draws 3 kW of active power and 4 kvars of reactive power. What effect does this
have on the reactive power Q provided by the ac power source and the power
factor cos Φ?
a.
b.
c.
d.
Q goes from +4 to -4 kvars and cos Φ is corrected to unity.
Q goes from +4 to -8 kvars and cos Φ remains the same.
Q goes from +4 to -4 kvars and cos Φ remains the same.
Q goes from +4 to -8 kvars and cos Φ is corrected to unity.
Instructor Approval: _____________________________________ Date: __________
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