Three-Phase Systems Review - SaskPower 3 Book 1/4. THREE-PHASE SYS… · Rationale: All utilities...

March 2, 2005

S T U D E N T M A N U A L

Three-Phase Systems Review

Copyright 2004 by the Training and Development Centre, SaskPower. All Rights Reserved

2 S T U D E N T T R A I N I N G M A N U A L

Prerequisites:

• None.

Objectives: From memory, you will be able to recall the connections, characteristics and load check values in a wye and delta system.

Rationale: All utilities have a large variety of three-phase voltages and services. A good understanding of the systems will prepare you for three-phase construction and maintenance.

Learning Objectives

• Explain the generation of a three-phase voltage.• Describe the connections used in a wye and delta system.• Explain the characteristics of a wye and delta connected system.• Calculate line and coil values for current and voltage in a wye and

delta system.• Calculate the power factor, phase angle, and load in kVA and kW in

a balanced wye and delta system.

Learning Methods

• Self-learning + On-the-job• Self-learning + On-the-job• Self-learning + On-the-job• Self-learning + On-the-job• Self-learning + On-the-job

EVALUATION METHODS

• Written test• Written test• Written test• Written test• Written test

T H R E E - P H A S E S Y S T E M S R E V I E W 3


STUDENT RESOURCES

• Delmar’s Standard Textbook of Electricity

Learning Steps

1. Read the Learning Guide.2. Follow the steps outlined in the Learning Guide.3. Clarify any questions or concerns you may have.4. Complete the Practice and Feedback.5. Complete the Evaluation.



Lesson 1: Generation of a Three-Phase VoltageLearning Objective:Explain the generation of a three-phase voltage.Learning Method:Self-learning + On-the-jobEvaluation Method:Written test

Introduction

In single-phase generation, a magnetic field cuts a coil of wire and avoltage is generated across the coil ends, as illustrated by the sine wavebelow.

By simply adding two more coils of wire to a single-phase generator,three distinct and separate voltages are produced.



The coils are 120 degrees apart, and the result is three identical electricalsine waves which provide uniform three-phase power.

---Note---In the above configuration, six line wires (service wires) areneeded to provide three-phase service.



Each of these three voltages reach their peak values at different times.Since the coils are spaced 120 degrees apart from each other, the peakvalue for each phase is reached 120 degrees apart from the other. Theorder in which phase voltage reaches peak value is called phasesequence or rotation. Positive phase rotation is red, yellow, blue or anycombination therein (RYBRYBRYBRYB).

There are two basic three-phase connections - wye or star connectionand delta connection.



Lesson 2: Wye and Delta ConnectionsLearning Objective:Describe the connections used in a wye and delta system.Learning Method:Self-learning + On-the-jobEvaluation Method:Written test

Wye Connections

In the wye or star configurations, like ends are connected to a commonpoint (star point). In some transformer configurations, it may benecessary to ground the star point.

Figure 1. Star Point

The wye connection provides three-phase power using three line wiresinstead of six. This results in less line loss. If the star point is grounded,there would be four wires to consider.

Delta Connections

The delta configuration requires that all unlike ends are connectedtogether.



The delta connection only requires three line wires, which will reduceline loss.



Lesson 3: Characteristics of a Wye or Delta SystemLearning Objective:Explain the characteristics of a wye and delta connected

system.Learning Method:Self-learning + On-the-jobEvaluation Method:Written test

Wye Systems

Wye systems have the following three characteristics:

• Like terminals connect at a common point to form a 3- or 4-wiresystem

• E line = E coil x 1.73• I line = I coil

Line voltage is measured across any two phases. Since these phases are120 degrees apart from each other, the coil voltages must be addedvectorally. In the wye system, the result is:

E line = E coil x 1.73

Delta Systems

Delta systems have the following three characteristics:

• Unlike terminal ends are connected together• E line = E coil

• I line = I coil x 1.73

Like the voltage line values in the wye system, the line current in thedelta system is established through vectoral addition as well. The resultis:

I line = I coil x 1.73

Reference

For more information on vectoral addition in wye and deltasystems, refer to Delmar’s Standard Textbook of Electricity.



Lesson 4: Line and Coil ValuesLearning Objective:Calculate line and coil values for current and voltage in a wye

and delta system.Learning Method:Self-learning + On-the-jobEvaluation Method:Written test

Introduction

Calculating line and coil values for E and I in a wye or delta system canbe accomplished by using the characteristics of both connections.

Example:

Calculate E coil and I coil for the generator and load configurationshown above.

First, calculate E coil of the load side.

In a wye system:

Therefore:



Since:

Now, calculate I coil of the load side. Since the impedance is given as32 ohms, Ohm’s Law is used to calculate I coil as follows:

The line voltage was given at 480V. The line amperage is determinedas follows:

In a wye system:

Therefore:

Now, calculate the E coil and I coil values of the generator.



We know, in a delta system:

We also know in a delta system:

Since:



Lesson 5: Power Factor, Phase Angle and LoadLearning Objective:Calculate the power factor, phase angle, and load in kVA and

kW in a balanced wye and delta system.Learning Method:Self-learning + On-the-jobEvaluation Method:Written test

Introduction

The power factor (p.f.) is the ratio of true power to apparent power.

The phase angle is the angular displacement by which current andvoltage are out of phase with each other.

The load in kVA is the apparent power.

The load in kW is the true power.

Any of the above variables can be calculated given the properinformation. Consider the following problem:

A three-phase generator, connected in wye, delivers power to a deltaconnected motor. The motor runs with a lagging power factor of 86%.Find three-phase kVA, three-phase kW and phase angle.

Calculate the motor load in kVA using a line values formula:



Calculate the motor load in kW:

Calculate the phase angle:

---Note---To make this calculation on a calculator, press .86, 2nd function,cos. In this case, the answer is 30.68 degrees lagging.



---Note---Remember, when doing any calculations in three-phase systems,check the requirements for wye or delta connections to arrive atthe correct coil and line values.



Summary

To summarize this module, you have learned:

• Three-phase generation.• Wye and delta connections.• Three-phase power factor, phase angle and load connections.

Practice Feedback

Review the lesson, ask any questions and complete the self test.

Evaluation

When you are ready, complete the final test. You are expected toachieve 100%.



Review Questions

1. The coils in a three-phase generator are:

(a) 90 degrees apart.

(b) 120 degrees apart.

(c) 180 degrees apart

(d) 360 degrees apart.

2. The two basic three-phase connections are:

(a) Wye and star.

(b) Wye and delta.

(c) Delta and wye-delta.

(d) None of these

3. Correct, positive phase rotation is:

(a) RYB.

(b) YBR.

(c) BRY.

(d) All of these

T / F 4. The order at which phase voltage reaches peak value is called phase peaking.

T / F 5. The order in which phase voltage reaches peak value is called phase sequence or rotation.

T / F 6. Three phase provides uniform power.

T / F 7. In the wye or star configuration, the same ends are connected to a common point.

T / F 8. It is necessary to always ground the star point.

T / F 9. The common point in a wye system, where like ends are connected, may also be called the star point.

T / F 10. The star configuration requires that all unlike ends are to be connected.

T / F 11. The delta configuration requires that all unlike ends are to be connected.



12. In a wye system, E line equals:

(a) E coil x I coil.

(b) E coil x I line.

(c) I coil x 1.73.

(d) E coil x 1.73.

13. In a wye system, I coil equals:

(a) I line.

(b) I line x 1.73.

(c) I line x E line.

(d) E line x 1.73.

14. In a delta system, E line equals:

(a) E coil x I coil.

(b) E coil.

(c) E coil x 1.73.

(d) I coil x 1.73.

15. In a delta system, I line equals:

(a) I coil.

(b) I coil x 1.73.

(c) E coil x 1.73.

(d) E coil x I coil.

T / F 16. Delta systems have unlike terminals connected together.

T / F 17. Wye systems have unlike terminals connected together.

T / F 18. Delta systems have like terminals connected together.

T / F 19. Wye systems have like terminals connected together.

20. From the graphic below the I coil of the source is:



(a) 6.28A.

(b) 10.84A.

(c) 62.28A.

(d) 108.4A.

21. From the graphic below the E coil of the source is:

(a) 277.42V.

(b) 347.21V.

(c) 373.68V.

(d) 646.46V.

22. From the graphic below the I line is:

(a) 6.28A.

(b) 10.84A.

(c) 62.28A.

(d) 108.4A.

23. From the graphic below the E line is:



(a) 277.42V.

(b) 347.21V.

(c) 373.68V.

(d) 646.46V.

24. From the graphic below the E coil of the load is:

(a) 62.28A.

(b) 347.21V.

(c) 373.68V.

(d) 646.46V.

25. From the graphic below the I line of the circuit is:

(a) 5.01A.

(b) 6.27A.

(c) 8.67A.

(d) 10.84A.

26. From the graphic below I coil of the source is:



(a) 5.01A.

(b) 6.27A.

(c) 8.67A.

(d) 10.84A.

27. From the graphic below E coil of the source is:

(a) 277V.

(b) 347V.

(c) 480V.

(d) 600V.

28. From the graphic below the I coil of the load is:

(a) 5.01A.

(b) 6.27A.

(c) 8.67A.

(d) 10.84A.

29. From the graphic below the E coil of the load is:



(a) 277V.

(b) 347V.

(c) 480V.

(d) 600V.

T / F 30. The phase angle is the angular displacement by which current and voltage are out of phase with each other.

31. The load in _____ is true power.

(a) kVA

(b) kW

(c) Volts

(d) Amps

32. The load in kVA is:

(a) True power.

(b) Apparent power.

(c) Phase angle.

(d) Reactive power.

T / F 33. The power factor is the ratio of true power to apparent power.

34. From the graphic below the apparent power of the generator is:

(a) 36.8kW.



(b) 36.8kVA.

(c) 40.44kW.

(d) 40.44kVA.

35. From the graphic below true power of the generator is:

(a) 36.8kW.

(b) 36.8kVA.

(c) 40.44kW.

(d) 40.44kVA.

36. From the graphic below the phase angle is:

(a) 21 degrees.

(b) 24.5 degrees.

(c) 31.7 degrees.

(d) 9.1 degrees.



Review Question Solutions

1. 120 degrees apart.

2. Wye and delta.

3. All of these

4. F

5. T

6. T

7. T

8. F

9. T

10. F

11. T

12. Ecoil x 1.73.

13. Iline.

14. Ecoil.

15. Icoil x 1.73.

16. T

17. F

18. F

19. T

20. 62.28A.

21. 373.68V.

22. 62.28A.



23. 646.46V.

24. 646.46V.

25. 10.84A.

26. 6.27A.

27. 600V.

28. 10.84A.

29. 347V.

30. T

31. kW

32. Apparent power.

33. T

34. 40.44kVA.

35. 36.8kW.

36. 24.5 degrees.

Three-Phase Systems Review - SaskPower 3 Book 1/4. THREE-PHASE SYS… · Rationale: All utilities...

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