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Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor.

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Presentation on theme: "Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor."— Presentation transcript:

1 Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor : IMS StafffCreation date : 01 Nov 2012Classification : D Transformers

2 Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor : IMS StafffCreation date : 01 Nov 2012Classification : D3 The intent of this presentation is to present enough information to provide the reader with a fundamental knowledge of transformers used within Michelin and to better understand basic system and equipment operations.

3 Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor : IMS StafffCreation date : 01 Nov 2012Classification : D3 Single Phase Transformers Ideal Transformer Characteristics The basic transformer consists of two coils electrically insulated from each other and wound upon a common core. Magnetic coupling is used to transfer electric energy from one coil to another. The coil, which receives energy from the AC source, is called the primary. The coil, which delivers energy to the AC load, is called the secondary. The core of transformers used at low frequencies is generally made of magnetic material, usually sheet steel. Cores of transformers used at higher frequencies are made of powdered iron and ceramics, or nonmagnetic materials. Some coils are simply wound on nonmagnetic hollow forms such as cardboard or plastic so that the core material is actually air Transformers

4 Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor : IMS StafffCreation date : 01 Nov 2012Classification : D3 Single Phase Transformers If a transformer is assumed to be operating under an ideal or perfect condition, the transfer of energy from one voltage to another is accompanied by no losses. Voltage Relationship The voltage (V) on the coils of a transformer is directly proportional to the number (N) of turns on the coils. This voltage relationship is expressed by the formula: Where:V pri = voltage on primary coil, V V sec = voltage on secondary coil, V N pri = number of turns on primary coil N sec = number of turns on secondary coil Vpri / Vsec = Npri / Nsec 07 - Transformers

5 Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor : IMS StafffCreation date : 01 Nov 2012Classification : D3 Single Phase Transformers Step-up and Step-down transformers A voltage ratio of 1:4 (read as 1 to 4) means that for each volt on the transformer primary, there is 4 volts on the secondary. When the secondary voltage is greater than the primary voltage, the transformer is a step-up transformer. A voltage ratio of 4:1 means that for every 4 volts on the primary, there is only 1 volt on the secondary. When the secondary voltage is less than the primary voltage, the transformer is called a step-down transformer. The voltage relationship discussed previously will apply to both step-up and step-down transformers. Current Relationship The current (I) in the coils of a transformer is inversely proportional to the voltage (V) on the coils. This current relationship is expressed by the formula: Where:V pri = voltage on primary coil, V V sec = voltage on secondary coil, V I pri = current in the primary coil, A I sec = current in the secondary coil, A 07 - Transformers

6 Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor : IMS StafffCreation date : 01 Nov 2012Classification : D3 Single Phase Transformers Efficiency The efficiency of a transformer is equal to the ratio of the power output of the secondary winding to power input of the primary winding. An ideal transformer is 100% efficient because it delivers all the energy it receives. Because of core and copper losses, the efficiency of even the best practical transformer is less than 100%. Efficiency expressed as an equation is: Where: Eff = efficiency P sec = power output from the secondary coil, W P pri = power input to the primary coil, W 07 - Transformers

7 Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor : IMS StafffCreation date : 01 Nov 2012Classification : D3 Single Phase Transformers Transformer Power Ratings Transformer capacity is rated in kilo-volt-amperes (KVA). This transformer rating is sometimes represented by the letter S. Since power in an AC circuit depends on the power factor of the load and the current in the load, an output rating in kilowatts must specify the power factor. For an ideal transformer we would assume efficiency to be 100%. Since our calculations will deal with ideal transformers, we can state the following relationships: Where: VA pri = power input to the primary coil, VA VA sec = power output from the secondary coil, VA V pri = primary voltage I pri = primary current V sec = secondary voltage I sec = secondary current 07 - Transformers

8 Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor : IMS StafffCreation date : 01 Nov 2012Classification : D3 Single Phase Transformers Dual Voltage Primary Wiring Diagram Low Voltage Connection 07 - Transformers

9 Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor : IMS StafffCreation date : 01 Nov 2012Classification : D3 Single Phase Transformers High Voltage Connection 07 - Transformers

10 Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor : IMS StafffCreation date : 01 Nov 2012Classification : D3 Single Phase Transformers Single Phase Transformer Calculations Given:N pri = 1500 turns N sec = 1000 turns V pri = 100 V Find:V sec = 07 - Transformers N sec N pri V V sec I pri I sec Solution: Npri / Nsec = Vpri / Vsec 1500 / 1000 = 100 / x X = Volts

11 Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor : IMS StafffCreation date : 01 Nov 2012Classification : D3 Single Phase Transformers Single Phase Transformer Calculations Given:N pri = 4000 turns N sec = 400 turns I sec = 6 A Find:I pri = What is the turn’s ratio? 07 - Transformers N sec N pri V V sec I pri I sec Solution: Npri / Nsec = turns ratio 4000 / 40 = 10 Np / Ns = Is / Ip 4000 / 400 = 6 / x x = 0.6 A

12 Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor : IMS StafffCreation date : 01 Nov 2012Classification : D3 Single Phase Transformer Calculations Given:V pri = 300 V V sec = 24 V I sec = 1 A N pri = 1000 turns Find:N sec = I pri = Solution: N sec = 80; I pri = 0.08 A 07 - Transformers N sec N pri V V sec I pri I sec Given:VA pri = 100 V pri = 460 V V sec = 110 V Find:I pri = I sec = Solution: I pri = A; I sec = A

13 Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor : IMS StafffCreation date : 01 Nov 2012Classification : D3 Single Phase Transformers Single Phase Transformer Calculations Given: N pri = 1000 turns N sec = 500 turns V pri = 200 volts Find:V sec Solution: V sec = 100 V 07 - Transformers Given: N pri = 5000 turns N sec = 500 turns I sec = 5 amps Find: I pri Solution: I pri = 0.5 A

14 Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor : IMS StafffCreation date : 01 Nov 2012Classification : D3 Single Phase Transformers Single Phase Transformer Calculations Given: N pri = 1000 turns V pri = 200 volts V sec = 24 volts I sec = 2 amps Find: N sec I pri 07 - Transformers Given: V sec = 110 volts S sec = 1100 VA N sec = 55 turns V pri = 200 volts Find: I sec; I pri; N pri; S pri Ratio of the transformer

15 Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor : IMS StafffCreation date : 01 Nov 2012Classification : D3 Single Phase Transformers Single Phase Transformer Calculations Given: V pri = 200 volts I pri = 5 amps N pri = 1000 turns N sec = 200 turns Find: V sec I sec 07 - Transformers

16 Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor : IMS StafffCreation date : 01 Nov 2012Classification : D3 Transformer Nameplate Data Introduction The data found on a typical transformer nameplate is shown below. Note that the transformer capacity is rated in KVA and not KW. The load connected to the secondary windings will determine the power factor, and thus, the apparent power for the output circuit of the transformer. Further, the KVA rating represents the full load output of the transformer and not the input. The other data on the nameplate usually includes the voltage rating of the high and low voltage windings, the frequency, number of phases, percent impedance, polarity, maximum temperature rise and the gallons of transformer oil required, if oil filled. Control Transformer Below is an example of a control transformer utilizing a dual voltage primary windings and a single voltage secondary winding. The primary voltage should be either 240 volts or 480 volts. The primary windings must be connected in parallel for 240 volts and connected in series for 480 volts. The secondary should always produce 24 volts Transformers

17 Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor : IMS StafffCreation date : 01 Nov 2012Classification : D3 Transformer Nameplate Data The power rating is 0.1 kVA or 100VA. Below is an example of a control transformer utilizing a multi-tap primary winding and a single voltage secondary winding. The primary voltage could be a variety of voltages ranging from 208 volts up to 480 volts. The primary windings must be connected according to the correct voltage applied to the primary winding. The secondary should always produce 24 volts Transformers

18 Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor : IMS StafffCreation date : 01 Nov 2012Classification : D3 Troubleshooting Single Phase Transformers The transformer below is a single phase control transformer connected for high voltage on the primary (480volts) and 120 volts on the secondary. To troubleshoot the transformer with an ohm-meter you should check the ohmic value for the primary and then the ohmic value for the secondary. You must make sure that the wires connecting the primary and secondary windings are not connected to any other device that would cause a reading through them. The ohmic valve will depend on the size of the transformer. The reading will usually be a low ohmic value which should show continuity through the windings Transformers

19 Presentation : IMS – Tech Managers ConferenceAuthor : IMS StaffCreation date : 08 March 2012Classification : D3Conservation :Page : ‹#› 07 - TransformersAuthor : IMS StafffCreation date : 01 Nov 2012Classification : D3 Troubleshooting Single Phase Transformers The transformer below is a single phase power transformer connected for high voltage on the primary (480volts) and 240 volts on the secondary. To troubleshoot the transformer with an ohm-meter you should check the ohmic value for the primary and then the ohmic value for the secondary. You must make sure that the wires connecting the primary and secondary windings are not connected to any other device that would cause a reading through them. The ohmic value will depend on the size of the transformer. The reading will usually be a low ohmic value which should show continuity through the windings Transformers


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