ECE 4501 Power Systems Laboratory Manual Rev 1.0

3.0 THREE-PHASE AC CIRCUITS

3.1 THREE-PHASE VOLTAGE AND CURRENT

3.1.1 OBJECTIVE

To study the relationship between voltage, current and power in three-phase circuits and examine wye and delta connections.

3.1.2 DISCUSSION

In practice three-phase circuits in proper working order tend to be symmetrical, consisting of three identical branches, each with the same impedance, voltage and current. Each of the branches can be treated as a single-phase circuit, and thus the analysis of a three-phase circuit is not necessarily any more difficult than the corresponding single-phase network. Unbalanced three-phase networks usually represent a problematic condition, such as a fault, and can be much more difficult to analyze. Unbalanced conditions are beyond the scope of this laboratory exercise.

Three-phase systems are normally configured in either a WYE or DELTA. The names result from the physical similarity of the wye connection to the letter Y and the delta connection to the Greek letter of the same name (a triangular symbol). Each of these configurations has unique electrical characteristics.

3.1.3 INSTRUMENTS AND COMPONENTS

Power Supply Module EMS 8821 Resistance Module EMS 8311 AC Metering Module (2.5 A) EMS 8425 AC Metering Module (250 V) EMS 8426

3.1.4 PROCEDURE

CAUTION! – High voltages are present in this Experiment. DO NOT make any connections with the power supply ON. Get in the habit of turning OFF the power supply after every measurement.

1) Use the power supply and AC voltmeter module to connect the following circuit:

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V 4 to 5 = ______V rms

V 5 to 6 = ______V rms

V 6 to 4 = ______V rms

V L-L Avg. = ______V rms

2) Turn on the power supply and adjust the voltage control knob until the line-to-neutral

voltage, V4N is exactly 120 V rms, as read from the power supply’s voltmeter.

3) Measure and record each Line-to-Line voltage from the voltmeters.

4) Turn the voltage control knob to zero percent and turn OFF the power supply.

5) Calculate the numeric average of the three line-to-line voltages and enter the result above.

6) Reconnect the voltmeters to read the line-to-neutral voltages of the power supply as shown below:

V 4 to N = ______V rms

V 5 to N = ______V rms

V 6 to N = ______V rms

V L-N Avg. = ______V rms

7) Turn on the power supply and adjust the voltage control knob until the line-to-neutral

voltage, V4N is exactly 120 V rms, as read from the power supply’s voltmeter.

8) Measure and record each Line-to-Neutral voltage (also called Phase Voltage) from the voltmeters.

10) Calculate the numeric average of the three line-to-neutral voltages and enter the result above.

11) Calculate the ratio of the average line-to-line voltage to the average line-to-neutral voltage and enter the result here:

______

12) Is this ratio approximately the square root of three? ______

13) Repeat the voltage measurement procedures for line-to-line and line-to-neutral voltages, steps 1 through 9 above, on the fixed 120/208 V power supply (terminals 1,2,3, and N) and enter the results below:

Fixed Supply: Line-to-line Fixed Supply: Line-to-Neutral

Line Voltages: Phase Voltages:

V 1 to 2 = ______V rms V 1 to N = ______V rms

14) Are the readings for each phase similar to the readings of the other two phases for each of the measurements? ______

15) Is the voltage between any two terminals, for example, V 3 to 1, V 5 to N or perhaps even V 3

to 5, a single-phase or a three-phase voltage? ______

16) Using the modules listed under Instruments and Components, build the following WYE circuit. DO NOT CONNECT THE NEUTRAL POINT OF THE WYE LOAD WITH THE NEUTRAL OF THE POWER SUPPLY.

Phase Voltages and Phase Currents:

V1 = ______V rms

V2 = ______V rms

V3 = ______V rms

I1 = ______A rms

I2 = ______A rms

I3 = ______A rms

17) Set each resistance section to 400 Ohms.

18) Turn on the power supply and adjust the voltage control until the meter on the power supply reads, V45 = 208 V rms (line-to-line).

19) Measure and record the voltages and currents on the instruments. Calculate the numeric average for both voltage and current for use in the table below.

21) Are the voltages and currents reasonably balanced? ______

22) Calculate the average phase voltage and current and also the apparent power (S = VI) dissipated by each phase and also the power delivered by the source: (Note: For a pure resistance, the apparent power, S, is equal to the real power, P.)

Average Value of Phase Voltage and Three Phase Power in a WYE: Phase Current:

Ptot = P1 + P2 + P3 = ______W Vphase = ______V Iphase = ______A

S3 = 3 VLineILine S1 = V1I1 = ______VA = P1

S2 = V2I2 = ______VA = P2 = 3 (3 Vphase)Iphase = ______VA

S3 = V3I3 = ______VA = P3 S3 = P3 = Ptot

23) Connect the following DELTA circuit:

Phase Voltages and Phase Currents:

V1 = ______V rms

V2 = ______V rms

V3 = ______V rms

I1 = ______A rms

I2 = ______A rms

I3 = ______A rms

24) Set each resistance section to 400 Ohms.

25) Turn on the power supply and adjust the voltage control until the meter on the power supply reads V45 = 120 V rms (line-to-line).

26) Measure and record the phase voltages and currents from the instruments.

27) Turn the voltage control knob to zero percent and turn OFF the power supply

28) Are the voltages and currents reasonably balanced? ______

29) Calculate the average phase voltage and current and also the apparent power ( S = VI) dissipated by the resistor in each phase:

Average Value of Phase Voltage and Phase Current: S1 = V1I1 = ______VA = P1

Vphase = ______V S2 = V2I2 = ______VA = P2

Iphase = ______A S3 = V3I3 = ______VA = P3

30) Disconnect the three current meters and insert them in series with the power supply terminals. Rebuild the DELTA by placing connection leads as needed such that the circuit looks like the one shown below:

Line Currents:

I4 = ______A

I5 = ______A

I6 = ______A

Iline = ______A

31) Turn on the power supply and adjust the voltage control until the meter on the power

supply reads V45 = 120 V rms (line-to-line).

32) Measure and record the line currents on the ammeters and calculate the numeric average,

Iline.

33) Turn the voltage control knob to zero percent and turn OFF the power supply

34) Calculate the ratio of the average line current, Iline, to the average phase current, Iphase. Also calculate the three-phase power delivered by the source:

Ratio of Line Current to Phase Three Phase Power in a DELTA: Current in a Delta Circuit: Ptot = P1 + P2 + P3 = ______W

Iline/Iphase = ______S3 = 3 Vphase Iphase = 3 VLine ILine Does this ratio equal the square root of three? ______= 3 Vphase (3Iphase)

= 3 Vphase ILine = ______VA

S3 = P3 = Ptot (S = P in purely resistive circuit, only)

3.1.5 CONCLUSIONS

1) In a WYE connected circuit, if the line-to-line voltage is 346 Volts, rms, what is the line-to-neutral voltage?

______

2) In a DELTA connected circuit, the current is 20 amperes in each phase. What is the line current?

______

3) In a WYE connected circuit the current is 10 Amperes per phase. What is the line current?

______

4) A WYE connected load has a resistance of 10 Ohms per phase. If the total three-phase power dissipated by the load is 3000 Watts, what is the line-to-line voltage of the source?

______

5) Three 11-Ohm resistors are connected in DELTA across a 3-phase, 440-Volt line. What is a) the line current and b) the total three-phase power?

______

3.2 THREE PHASE WATTS, VARS AND VOLT- AMPERES

3.2.1 OBJECTIVE

To determine the apparent, real and reactive power in three phase circuits and calculate power factor.

3.2.2 DISCUSSION

Section 3.1 examined the nature of real power in three phase circuits. By measuring the voltage drop across a load resistor and measuring the current flowing through it, one was able to calculate the apparent power dissipated by the resistor (S=VI). Since the load was a pure resistance, the apparent power, S, was equal to the real power, P. Mathematically, one would say that P = VI cos . The power factor angle, , is the phase angle difference between the voltage phasor and the current phasor, as determined by the equation,  = angle (V) – angle (I). In the case of a pure resistance, the voltage and current phasors are ‘in phase’ and thus  is zero degrees. Therefore, P = VI cos (0) = VI = S.

This section will demonstrate that reactive power, Q, either capacitive or inductive, can also be measured in the same manner. When the magnitude of the voltage drop across an inductor or capacitor is measured, along with the current flowing through it, one is able to say mathematically, Q = VI sin . In the case of a pure inductance, the voltage and current phasors are out of phase by 90 (The convention is to say that, for an inductor, the current lags the voltage by 90). Thus, Q = VI sin  = VI sin (90) = VI = S. However, in the case of a pure capacitance, the current phasor leads the voltage phasor by 90 and thus they are out of phase by -90. Hence, Q = VI sin (-90) = -VI = -S. It is therefore extremely important to know if the reactive load is capacitive or inductive because it will change the sign of the reactive power.

Knowing both real and reactive power in a circuit with a load consisting of both real and reactive elements, it is then possible to calculate the apparent power and power factor for balanced three phase circuits.

3.2.3 INSTRUMENTS AND COMPONENTS

Power Supply Module (0 – 120/208 Vac) EMS 8821 Resistance Module EMS 8311 Inductance Module EMS 8321 AC Metering Module (0.5 A, 3 phase) EMS 8425 AC Metering Module (250 V, 3 phase) EMS 8426

3.2.4 PROCEDURE

CAUTION! – High voltages are present in this Experiment. DO NOT make any connections with the power supply ON. Get in the habit of turning OFF the power supply after every measurement.

1) Using the equipment listed above, connect the WYE circuit shown below. DO NOT CONNECT THE NEUTRAL POINT OF THE THREE LOAD INDUCTORS TO THE NEUTRAL OF THE POWER SUPPLY.

Phase Voltages and Currents:

V1 = ______V rms

V2 = ______V rms

V3 = ______V rms

I1 = ______A rms

I2 = ______A rms

I3 = ______A rms

Average Values:

Vphase = ______V rms

Iphase = ______A rms

2) Set each inductance section to a reactance of 300 Ohms.

3) Turn on the power supply and adjust the voltage control for 208 V rms (line to line).

4) Measure and record the voltages and currents associated with the three inductive loads.

5) Return the voltage control to zero percent and turn OFF the power supply.

6) Calculate the numeric average of the phase currents and phase voltages and enter the results in the space provided above.

7) Calculate the apparent power for each of the inductive loads using S = VI. Note that since the loads are almost purely inductive, the apparent power, S, will be equal to the reactive power, Q. (See 3.2.2 DISCUSSION) Also calculate the total three-phase reactive power by summing the three individual phases.

(see table on next page)

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Apparent Power in Inductive Load: Three-phase Power:

V1I1 = ______VA = Q1 Q3 = Q1+ Q2+ Q3= ______VAR

V2I2 = ______VA = Q2 S3 = 3 VlineIline

V3I3 = ______VA = Q3 = 3 VphaseIphase = ______VA

8) Now calculate the three-phase apparent power, S3 = 3 VI and compare the result to the three phase sum calculated in 7) above. Do they compare favorably? ______

9) Using the Resistance Module, add 400 Ohms of resistors to each phase of the WYE load, in series with the inductors as shown below. DO NOT CONNECT THE NEUTRAL POINT OF THE LOAD TO THE NEUTRAL OF THE POWER SUPPLY.

Inductor Voltages and Phase Currents:

V1 = ______V rms

V2 = ______V rms

V3 = ______V rms

I1 = ______A rms

I2 = ______A rms

I3 = ______A rms

10) With the resistance set at 400 Ohms per phase and the inductance set at 300 Ohms per phase, turn on the power supply and adjust the voltage control for 208 V rms (line to line).

11) Measure and record the voltages and currents associated with the three inductive portions of the load.

13) Now reconnect the voltmeters so that they read the voltage drops associated with the resistors in the circuit, as shown in the diagram below:

Resistor Voltages:

V4 = ______V rms

V5 = ______V rms

V6 = ______V rms

14) Turn on the power supply and adjust the voltage control for 208 V rms (line to line).

15) Measure and record the voltage drops across the resistors in the circuit.

17) By measuring the phase current and the voltage drop across any component in a WYE load it is possible to calculate the apparent power dissipated in that component using S = VI. In the case of the inductive element, the apparent power will be purely reactive, Q (in VARs). In the case of a resistive component, the apparent power will be purely real, P. Now, using the voltages and currents measured in the last two circuits, calculate the real and reactive powers associated with the WYE load:

Real Power in the Resistors: Reactive Power in the Inductors:

P1 = I1 V4 = ______Watts Q1 = I1 V1 = ______VARs

P3 = P1 + P2 + P3 = ______W Q3 = Q1+ Q2+ Q3= ______VAR

18) Now calculate the total three-phase apparent power using the power triangle and the Pythagorean Theorem :

(see diagram on next page)

Apparent Power for Complex Load:

2 2 S3 = { (P3) + (Q3) }

S3 = ______VA

Which should be similar to:

S3 = 3 VlineIline = ______VA

19) Also calculate the total three-phase apparent power using S3 = 3 VlineIline :

20) Finally, calculate the power factor using the equation, pf = P/S: Power Factor, pf = ______Watts / ______VA = ______,

leading or lagging (circle one)

3.2.5 CONCLUSIONS

1) A three-phase motor draws 10 amps of current from a 440-volt line at a power factor of 80 percent, lagging. Calculate the three-phase apparent power, S, and the three-phase real power, P:

______

2) A transformer supplies 120 kVA to a three-phase load at a line voltage of 2400 V rms. Find the line current:

______