Single-Phase Transformation Review

Transcription

1 Single-Phase Transformation Review S T U D E N T M A N U A L March 2, 2005

2 2 STUDENT TRAINING MANUAL 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

3 SINGLE-PHASE TRANSFORMATION REVIEW 3 STUDENT RESOURCES Learning Steps Level 2 Training Manuals 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.

4 4 STUDENT TRAINING MANUAL Lesson 1: Single-Phase Transformer Operating Principles Learning Objective:Describe the operating principles of a single-phase transformer. Learning Method:Self-learning + On-the-job Evaluation Method:Written test Introduction In order for one to connect and operate a single-phase transformer, an understanding of the operating principles must first be undertaken. Law of Electromagnetic Induction The law of electromagnet induction states that whenever a conductor cuts through magnetic flux lines, a voltage is inducted into the conductor. Three factors affect the amount of induced voltage: Number of turns of wire Strength of the magnetic field Speed of the cutting action

5 SINGLE-PHASE TRANSFORMATION REVIEW 5 If a coil of wire is wrapped around a steel core and energized, the steel core becomes magnetized. Lines of force or flux lines from the magnet will be established and will expand and contract, cutting a second coil of wire. When these flux lines cut the secondary coil, a voltage is induced. Voltages can be stepped up or down, depending on the winding application. Purpose of Transformation The purpose of transformation is to raise or lower voltage by transferring electrical energy from one AC circuit to another. This transforming ability allows electrical energy to be used in a wide variety of applications (ie: running electrical motors, lighting and heating). Transformer Losses Although transformers tend to be between 90% to 99% efficient, small losses 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 the coils. 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 the core every 1/120 of a second.

6 6 STUDENT TRAINING MANUAL Eddy Current Loss Excitation Current Eddy current loss is caused by circular induced current opposite of the main currents. This opposition causes heat. Laminating the core will protect the transformer from eddy currents. Excitation current or charging current as it is often called, is the amount of current required to magnetize the core of a transformer. From no load to full load, the charging current remains constant. As a rule, excitation current is so small it is often ignored in kva load calculations.

7 SINGLE-PHASE TRANSFORMATION REVIEW 7 Lesson 2: Single-Phase Transformer Ratios Learning Objective:Calculate unknown voltage and current variables on a singlephase transformer. Learning Method:Self-learning + On-the-job Evaluation Method:Written test Introduction The transformer ratio is the relationship between the number of wraps of wire in the primary and secondary coil, and is directly related to primary and secondary voltage and inversely related to the primary and secondary current. If a tap changer is involved, incorporate it into the turns ratio immediately. Doing this will ensure a correct ratio for all calculations being done. Example: Tap changer set at 97.5%

8 8 STUDENT TRAINING MANUAL Lesson 3: Connection of Two Single-Phase Transformers in Parallel Learning 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-job Evaluation Method:Written test Selecting a Transformer to be Connected in Parallel When selecting a transformer to be connected in parallel, the following considerations 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.

9 SINGLE-PHASE TRANSFORMATION REVIEW 9 Lesson 4: Procedure to Calculate kva Learning Objective:Explain the procedure to calculate the load in kva on a singlephase transformer. Learning Method:Self-learning + On-the-job Evaluation Method:Written test Introduction Transformer load checks are calculated in kva and are obtained to prevent 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 measured with a clip-on ammeter, and the current readings are averaged.

10 10 STUDENT TRAINING MANUAL ---Note--- When performing load checks, the max-amp meter ratio must be considered and also the tap changer position.

11 SINGLE-PHASE TRANSFORMATION REVIEW 11 Find the watts on the circuit and the load in kva. Load calculations:

12 12 STUDENT TRAINING MANUAL

13 SINGLE-PHASE TRANSFORMATION REVIEW 13 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.

14 14 STUDENT TRAINING MANUAL Lesson 5: 3-Wire Service Voltage Calculations Learning Objective:Calculate the voltage available to the load in a 3-wire service with and without a neutral. Learning Method:Self-learning + On-the-job Evaluation Method:Written test Introduction All new homes are provided with a 120/240 volt 3-wire service - two hot leg wires and a neutral wire. The hot legs provide 120 volts phase-toground or 240 volts phase-to-phase to loads, and the neutral carries the amperage imbalance back to the source, as well as provides a degree of protection against damaged or removed insulation. To understand how important a neutral wire is to a service, the following load calculations are done. Unbalanced 3-Wire Circuits In an unbalanced single-phase 3-wire circuit, the current flowing through each hot leg is different. The imbalance or difference between the two is carried through the neutral wire. The neutral wire shares the

15 SINGLE-PHASE TRANSFORMATION REVIEW 15 current of the larger load. As illustrated above, R 1 carries the same current value as Load A, and R 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 A and Load B. Now we can calculate the voltage drops across the resistance of the line wires and the neutral wire.

16 16 STUDENT TRAINING MANUAL From the above values, we can solve for the voltages available to Loads A and B. Solve for the largest load first. Since Load B shares its current with the neutral wire, the voltage available to Load B must be equal to the voltage applied across the hot leg and the neutral (120V) with the line wire and neutral wire voltage losses subtracted. Therefore, we can calculate the voltage at Load B:

17 SINGLE-PHASE TRANSFORMATION REVIEW 17 The voltage available to Load A must be equal to the total voltage applied to the 3-wire circuit, which is now drawn as a 2-wire series circuit. The voltage at Load A will be equal to 240V minus E N, E 1 and E B. We know this is correct because the total applied voltage is equal 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 calculate the resistance for each load as follows:

18 18 STUDENT TRAINING MANUAL Broken Neutral in a 3-Wire Circuit There would be no ill effects if the neutral wire were to break in a balanced 3-wire circuit. However, if the circuit was unbalanced, one side of the circuit could receive more than its share of the applied voltage. In fact, in an extremely unbalanced circuit, one hot leg could receive nearly all of the applied voltage. This would, in turn, allow voltage way above our standard rated limits and cause damage to appliances and lights. Let s consider the possibility of a broken neutral in the previous example.

19 SINGLE-PHASE TRANSFORMATION REVIEW 19 What happens if the neutral wire breaks? What would be the applied voltage at Loads A and B? First of all, we should calculate the total resistance of the circuit:

20 20 STUDENT TRAINING MANUAL 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 neutral absent, 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.

21 SINGLE-PHASE TRANSFORMATION REVIEW 21 As you can see above, the voltages present at each load would cause considerable damage to appliances normally designed to operate on 120 volts. ---Note--- In all broken neutral cases, the load with the greater resistance receives the highest voltage.

22 22 STUDENT TRAINING MANUAL 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 singlephase 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 to achieve 100%.

23 SINGLE-PHASE TRANSFORMATION REVIEW 23 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. T / F 6. T / F 7. T / F 8. T / F 9. Excitation current or charging current is the amount of current required to magnetize the core of a transformer. Excitation current or charging current is the amount of current required to magnetize a coil. Excitation current is always figured into kva load calculations. Excitation current is so small it is often ignored in kva load calculations. The transformer ratio is the relationship between the number of wraps of wire in the primary and secondary coil.

24 24 STUDENT TRAINING MANUAL 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.

25 SINGLE-PHASE TRANSFORMATION REVIEW 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) (b) 71.4kVA kVA (c) kVA (d) 432kVA

26 26 STUDENT TRAINING MANUAL 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) (b) 64.26kW kW (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

27 SINGLE-PHASE TRANSFORMATION REVIEW What are the voltages available to the loads in the circuit below? (a) E A = volts, E B = 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 = 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

28 28 STUDENT TRAINING MANUAL 25. What is the voltage at Loads A and B when the neutral is broken? (a) E A = volts, E B = volts (b) E A = volts, E B = volts (c) E A = 120 volts, E B = 120 volts (d) E A = 240 volts, E B = 240 volts

29 SINGLE-PHASE TRANSFORMATION REVIEW 29 Review Question Solutions 1. T 2. The speed of the cutting action, strength of the magnetic field, and number of turns of wire. 3. T 4. Eddy current and hysteresis. 5. T 6. F 7. F 8. T 9. T 10. T 11. F 12. T 13. T 14. E P = 14,400 volts. 15. I P = 0.83 amps. 16. T :1 18. E P = 2380 volts 19. I average = 300 amps kVA kW

30 30 STUDENT TRAINING MANUAL 22. E 1 = 2.25 volts, E 2 = 6.75 volts, E N = 4.5 volts 23. E A = volts, E B = volts 24. R A = ohms, R B = 12.1 ohms 25. E A = volts, E B = volts