single-phase transformation review - saskpower 3 book 1/3. single...troubleshooting of single-phase...

March 2, 2005

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

Single-Phase Transformation

Review

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: Given the Construction Standards manual and a formula sheet, you will be able to explain transformation operating principles, load checks and paralleling two transformers.

Rationale: This review will assist you in the construction, maintenance and troubleshooting of single-phase and three-phase transformer connections.

Learning Objectives

• Describe the operating principles of a single-phase transformer.• Calculate unknown voltage and current variables on a single-phase

transformer.• Explain the procedure to select and connect two single-phase

transformers in parallel to provide a single-phase 3-wire service.• Explain the procedure to calculate the load in kVA on a single-phase

transformer.• Calculate the voltage available to the load in a 3-wire service with

and without a neutral.

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

S I N G L E - P H A S E T R A N S F O R M A T I O N R E V I E W 3

STUDENT RESOURCES

• Level 2 Training Manuals

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: Single-Phase Transformer Operating PrinciplesLearning Objective:Describe the operating principles of a single-phase transformer.Learning Method:Self-learning + On-the-jobEvaluation Method:Written test

Introduction

In order for one to connect and operate a single-phase transformer, anunderstanding of the operating principles must first be undertaken.

Law of Electromagnetic Induction

The law of electromagnet induction states that whenever a conductorcuts through magnetic flux lines, a voltage is inducted into theconductor.

Three factors affect the amount of induced voltage:

• Number of turns of wire• Strength of the magnetic field• Speed of the cutting action

If a coil of wire is wrapped around a steel core and energized, the steelcore becomes magnetized.

Lines of force or flux lines from the magnet will be established and willexpand and contract, cutting a second coil of wire. When these fluxlines cut the secondary coil, a voltage is induced.

Voltages can be stepped up or down, depending on the windingapplication.

Purpose of Transformation

The purpose of transformation is to raise or lower voltage bytransferring electrical energy from one AC circuit to another. Thistransforming ability allows electrical energy to be used in a wide varietyof applications (ie: running electrical motors, lighting and heating).

Transformer Losses

Although transformers tend to be between 90% to 99% efficient, smalllosses do occur during their regular operation.

There are two main types of transformer losses:

• Copper loss• Core loss

Copper Loss

Copper loss occurs when heat is produced by current flow through thecoils.

Core Loss

Two types of core loss are:

• Hysteresis loss• Eddy current loss

Hysteresis Loss

Hysteresis loss is caused by the molecules changing directions in thecore every 1/120 of a second.

Eddy Current Loss

Eddy current loss is caused by circular induced current opposite of themain currents. This opposition causes heat. Laminating the core willprotect the transformer from eddy currents.

Excitation Current

Excitation current or charging current as it is often called, is the amountof current required to magnetize the core of a transformer. From no loadto full load, the charging current remains constant. As a rule, excitationcurrent is so small it is often ignored in kVA load calculations.

Lesson 2: Single-Phase Transformer RatiosLearning Objective:Calculate unknown voltage and current variables on a single-

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

Introduction

The transformer ratio is the relationship between the number of wrapsof wire in the primary and secondary coil, and is directly related toprimary and secondary voltage and inversely related to the primary andsecondary current.

If a tap changer is involved, incorporate it into the turns ratioimmediately. Doing this will ensure a correct ratio for all calculationsbeing done.

Example:

Tap changer set at 97.5%

Lesson 3: Connection of Two Single-Phase Transformers in ParallelLearning Objective:Explain the procedure to select and connect two single-phase

transformers in parallel to provide a single-phase 3-wire service.Learning Method:Self-learning + On-the-jobEvaluation Method:Written test

Selecting a Transformer to be Connected in Parallel

When selecting a transformer to be connected in parallel, the followingconsiderations must be observed:

• The transformers must have identical high and low voltage ratings.• The transformer’s IZ values must be within +/- 75% of each other.• The transformer’s polarity (subtractive or additive) must be

correctly identified (nameplate, polarity check).• The primary source must be identically connected.

Connecting Two Transformers in Parallel

When connecting two transformers in parallel on a 3-wire service,complete the following steps:

• Inform the customer of the outage and its approximate duration.• Switch the circuit breaker to the “Off” position and remove the

meter, if applicable.• Isolate and de-energize the transformer from its primary source.• Install the transformer and connect the like polarity terminals.• Double check the connections (ensure maximum climbing space).• Energize the transformers and measure the secondary voltage to

ensure it is acceptable.• Re-install the meter and switch the circuit breaker to the “On”

position, if applicable.

Lesson 4: Procedure to Calculate kVALearning Objective:Explain the procedure to calculate the load in kVA on a single-

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

Introduction

Transformer load checks are calculated in kVA and are obtained toprevent overload of electrical circuits and connected apparatus.

Load Check Procedure for a Secondary Service

The procedure to perform a load check is as follows:

• Check the operation of the meter.• Determine if rubber gloves are required (300V to 5,000V).• Measure the current and voltage values using a voltmeter and

ammeter.• Average the current on the A and B phase.• Calculate using the kVA formula.

Load Check Formulas

Voltage is measured line to line with a voltmeter. Current is measuredwith a clip-on ammeter, and the current readings are averaged.

---Note---When performing load checks, the max-amp meter ratio must beconsidered and also the tap changer position.

Find the watts on the circuit and the load in kVA.

Load calculations:

Some things to remember about these examples:

• Load formula is for kVA readings.• Watts is simply true power.• Load in kVA is simply apparent power.

• Solve vectorally.

Lesson 5: 3-Wire Service Voltage CalculationsLearning Objective:Calculate the voltage available to the load in a 3-wire service

with and without a neutral.Learning Method:Self-learning + On-the-jobEvaluation Method:Written test

Introduction

All new homes are provided with a 120/240 volt 3-wire service - two hotleg wires and a neutral wire. The hot legs provide 120 volts phase-to-ground or 240 volts phase-to-phase to loads, and the neutral carries theamperage imbalance back to the source, as well as provides a degree ofprotection against damaged or removed insulation.

To understand how important a neutral wire is to a service, the followingload calculations are done.

Unbalanced 3-Wire Circuits

In an unbalanced single-phase 3-wire circuit, the current flowingthrough each hot leg is different. The imbalance or difference betweenthe two is carried through the neutral wire. The neutral wire shares the

current of the larger load.

As illustrated above, R 1 carries the same current value as Load A, andR 2 carries the same current value as Load B. The neutral wire, R N,carries the difference in the current values of Load A and Load B.

Therefore:

R N has 6A flowing through it. 6A is the difference between Load Aand Load B.

Now we can calculate the voltage drops across the resistance of the linewires and the neutral wire.

From the above values, we can solve for the voltages available to LoadsA and B.

Solve for the largest load first.

Since Load B shares its current with the neutral wire, the voltageavailable to Load B must be equal to the voltage applied across the hotleg and the neutral (120V) with the line wire and neutral wire voltagelosses subtracted.

Therefore, we can calculate the voltage at Load B:

The voltage available to Load A must be equal to the total voltageapplied to the 3-wire circuit, which is now drawn as a 2-wire seriescircuit. The voltage at Load A will be equal to 240V minus E N , E 1and E B. We know this is correct because the total applied voltage isequal to the sum of the voltage drops in the entire circuit.

Calculating E A :

With the E and I values of Loads A and B now known, we can calculatethe resistance for each load as follows:

Broken Neutral in a 3-Wire Circuit

There would be no ill effects if the neutral wire were to break in abalanced 3-wire circuit.

However, if the circuit was unbalanced, one side of the circuit couldreceive more than its share of the applied voltage. In fact, in anextremely unbalanced circuit, one hot leg could receive nearly all of theapplied voltage. This would, in turn, allow voltage way above ourstandard rated limits and cause damage to appliances and lights.

Let’s consider the possibility of a broken neutral in the previousexample.

What happens if the neutral wire breaks? What would be the appliedvoltage at Loads A and B?

First of all, we should calculate the total resistance of the circuit:

Now that R T is known, we can calculate I T using Ohm’s Law.

The current is 3.15A in each resistor due to the fact that, with the neutralabsent, we have a series circuit. In a series circuit, I remains constant.

Using Ohm’s Law, we can calculate the voltage available to each load.

As you can see above, the voltages present at each load would causeconsiderable damage to appliances normally designed to operate on 120volts.

---Note---In all broken neutral cases, the load with the greater resistancereceives the highest voltage.

Summary

To summarize this module, you have learned:

• The operating principles of a single-phase transformer.• How to calculate unknown voltage and current variables on a single-

phase transformer.• The procedure to select and connect two single-phase transformers

in parallel to provide a single-phase 3-wire service.• The procedure to calculate the load in kVA on a single-phase

transformer.• How to calculate the voltage available to the load in a 3-wire service

with and without a neutral.

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

T / F 1. The law of electromagnetic induction states that, whenever a conductor cuts through magnetic flux lines, a voltage is induced into the conductor.

2. Factors which affect the amount of induced voltage are:

(a) Primary voltage, strength of magnetic field, and breaker size.

(b) Strength of magnetic field, the speed of the cutting action, and breaker size.

(c) Number of turns of wire, the speed of the cutting action, and primary voltage.

(d) The speed of the cutting action, strength of the magnetic field, and number of turns of wire.

T / F 3. The purpose of transformation is to raise or lower the voltage by transferring electrical energy from one AC circuit to another.

4. Two types of core loss are:

(a) Copper and eddy current.

(b) Excitation current and hysteresis.

(c) Copper and excitation current.

(d) Eddy current and hysteresis.

T / F 5. Excitation current or charging current is the amount of current required to magnetize the core of a transformer.

T / F 6. Excitation current or charging current is the amount of current required to magnetize a coil.

T / F 7. Excitation current is always figured into kVA load calculations.

T / F 8. Excitation current is so small it is often ignored in kVA load calculations.

T / F 9. The transformer ratio is the relationship between the number of wraps of wire in the primary and secondary coil.

T / F 10. The transformer ratio is inversely related to the primary and secondary current.

T / F 11. In order to parallel transformers, their IZ values must be within +/- 10.5% of each other.

T / F 12. In order to parallel transformers, their high and low voltage ratings must be identical.

T / F 13. In order to parallel transformers, they must be connected to the same primary source.

14. The number of wire turns in a primary coil is 360 and the number of wire turns in a secondary coil is 6. The secondary voltage is 240 volts and the secondary current is 50 amps. The primary voltage is:

(a) E P = 14,400 volts.

(b) E P = 15,600 volts.

(c) E P = 1,440 volts.

(d) E P = 13,200 volts.

15. The number of wire turns in a primary coil is 360 and the number of wire turns in the secondary coil is 6. The secondary voltage is 240 volts and the secondary current is 50 amps. The primary current is:

(a) I P = 0.83 amps.

(b) I P = 8.3 amps.

(c) I P = 300 amps.

(d) I P = 0.3 amps.

T / F 16. When performing a load check, an operator may require the use of a voltmeter, ammeter and rubber gloves.

17. A transformer is rated at 2,400V - 120V/240V. The secondary voltage is 238V. Two max-amp meters are connected to the secondary hot legs on an 8:1 ratio. The readings are: A = 33A, B = 42A. The PF is 90%. Calculate the transformer ratio.

(a) 10:1

(b) 20:1

(c) 8:1

(d) 5:1

18. A transformer is rated at 2,400V - 120V/240V. The secondary voltage is 238V. Two max-amp meters are connected to the secondary hot legs on an 8:1 ratio. The readings are: A = 33A, B = 42A. The PF is 90%. Calculate the actual primary voltage.

(a) E P = 2380 volts

(b) E P = 2400 volts

(c) E P = 1920 volts

(d) E P = 2160 volts

19. A transformer is rated at 2,400V - 120V/240V. The secondary voltage is 238V. Two max-amp meters are connected to the secondary hot legs on an 8:1 ratio. The readings are: A = 33A, B = 42A. The PF is 90%. The actual average secondary amperage is:

(a) I average = 300 amps.

(b) I average = 54 amps.

(c) I average = 240 amps.

(d) I average = 540amps.

20. A transformer is rated at 2,400V - 120V/240V. The secondary voltage is 238V. Two max-amp meters are connected to the secondary hot legs on an 8:1 ratio. The readings are: A = 33A, B = 42A. The PF is 90%. Calculate the load in kVA.

(a) 71.4kVA

(b) 12.852kVA

(c) 12.960kVA

(d) 432kVA

21. A transformer is rated at 2,400V - 120V/240V. The secondary voltage is 238V. Two max-amp meters are connected to the secondary hot legs on an 8:1 ratio. The readings are: A = 33A, B = 42A. The PF is 90%. Calculate the true power in the circuit.

(a) 64.26kW

(b) 11.571kW

(c) 11.66kW

(d) 38.88kW

22. What is the voltage drop in the wires in the circuit below?

(a) E 1 = 2.25 volts, E 2 = 6.75 volts, E N = 4.5 volts

(b) E 1 = 120 volts, E 2 = 120 volts, E N = 240 volts

(c) E 1 = 4.5 volts, E 2 = 4.5 volts, E N = 4.5 volts

(d) E 1 = .75 volts, E 2 = .75 volts, E N = .75 volts

23. What are the voltages available to the loads in the circuit below?

(a) E A = 122.25 volts, E B = 108.75 volts

(b) E A = 120 volts, E B = 120 volts

(c) E A = 240 volts, E B = 240 volts

(d) E A = 2.25 volts, E B = 6.75 volts

24. What are the resistances of Load A and B in the circuit below?

(a) R A = 40.75 ohms, R B = 12.1 ohms

(b) R A = 3 ohms, R B = 9 ohms

(c) R A = 3 amps, R B = 9 amps

(d) R A = 44.7 ohms, R B = 75.9 ohms

25. What is the voltage at Loads A and B when the neutral is broken?

(a) E A = 179.3 volts, E B = 53.24 volts

(b) E A = 108.75 volts, E B = 122.25 volts

(c) E A = 120 volts, E B = 120 volts

(d) E A = 240 volts, E B = 240 volts

Review Question Solutions

2. The speed of the cutting action, strength of the magneticfield, and number of turns of wire.

4. Eddy current and hysteresis.

14. EP = 14,400 volts.

15. IP = 0.83 amps.

17. 10:1

18. EP = 2380 volts

19. Iaverage = 300 amps.

20. 71.4kVA

21. 64.26kW

22. E1 = 2.25 volts, E2 = 6.75 volts, EN = 4.5 volts

23. EA = 122.25 volts, EB = 108.75 volts

24. RA = 40.75 ohms, RB = 12.1 ohms

25. EA = 179.3 volts, EB = 53.24 volts

single-phase transformation review - saskpower 3 book 1/3. single...troubleshooting of single-phase...

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