Showing posts with label Tutorials. Show all posts
Showing posts with label Tutorials. Show all posts

POWER TRANSFORMER QUESTION AND ANSWER TUTORIALS PART 1

QUESTION AND ANSWER ABOUT POWER TRANSFORMER
Power Transformer Question and Answer

1 . What is a transformer and how does it work?

A transformer is an electrical apparatus designed to convert alternating current from one voltage to another. It can be designed to “step up” or “step down” voltages and works on the magnetic induction principle.

A transformer has no moving parts and is a completely static solid state device, which insures, under
n o rmal operating conditions, a long and t ro u b l e - f ree life. It consists, in its simplest form, of two or more coils of insulated wire wound on a laminated steel core.

When voltage is introduced to one coil, called the primary, it magnetizes the iron core . A voltage is then induced in the other coil, called the secondary or output coil. The change of voltage (or voltage ratio) between the primary and secondary depends on the turns ratio of the two coils.

2 .What are taps and when are they used?
Taps are provided on some transformers on the high voltage winding to correct for high or low voltage conditions, and still deliver full rated output voltages at the secondary terminals. Standard tap arrangements are at two-and-one -half and five percent of the rated primary voltage for both high and low voltage conditions.

For example, if the  transformer has a 480 volt primary and the available line voltage is running at 504 volts, the primary should be connected to the 5% tap above normal in order that the secondary voltage be maintained at the proper rating.

The standard ASA and NEMA designation for taps are “ANFC” (above normal full capacity) and “B N FC” (below normal full capacity).


3 . What is the difference between “ I n s u l a t i n g,” “I s o l a t i n g,” and “Shielded Winding” transformers?
Insulating and isolating transformers are identical. These terms are used to describe the isolation of the primary and secondary windings, or insulation between the two.

A shielded transformer is designed with a metallic shield between the primary and secondary windings to attenuate transient noise. This is especially important in critical applications such as computers, process controllers and many other microprocessor controlled devices.

All two, three and four winding transformers are of the insulating or isolating types. Only autotransformers ,
w hose primary and secondary are connected to each other electrically, are not of the insulating or isolating variety.

4. Can transformers be operated at voltages other than nameplate voltages?
In some cases, transformers can be operated at voltages below the nameplate rated voltage. In N O case should a transformer be operated at a voltage in excess of its nameplate rating unless taps are provided for this purpose.

When operating below the rated voltage, the K VA capacity is reduced correspondingly. For example, if a 480 volt primary transformer with a 240 volt secondary is operated at 240 volts, the secondary voltage is reduced to 120 volts.

If the transformer was originally rated 10 KVA, the reduced rating would be 5 KVA, or in direct proportion to the applied voltage.

5. Can 60 Hz transformers be operated at 50 Hz?
ACME transformers rated below 1 KVA c a n be used on 50 Hz service. Transformers 1 KVA and larger, rated at 60 Hz, should not be used on 50 Hz service due to the higher losses and resultant heat rise. Special designs are required for this service.

However, any 50 Hz transformer will operate on a 60 Hz service .

SIMPLE POWER TRANSFORMER CONSTRUCTION BASIC TUTORIALS

POWER TRANSFORMER OPERATION SIMPLE ILLUSTRATION
How Power Transformer Works?

From the basic principles of magnetic induction, it is not difficult to see how a rudimentary transformer could be constructed. If a conductor carrying a changing current is brought near a second conductor, then the changing magnetic flux surrounding the first conductor will be linked to the second conductor and will induce a voltage.

Such a rudimentary transformer is depicted in Figure 1.3.
An AC voltage is connected to a primary conductor, shown as the left hand solid bar in Figure 1.3. In response to the voltage, an AC current flows, setting up a time-varying magnetic field surrounding the primary conductor.

A secondary conductor, shown as the right-hand solid bar, is located in proximity to the primary conductor so that the magnetic flux surrounding the primary conductor links the secondary circuit. According to the law of induction, there will be an induced voltage E around the path surrounding the time varying flux.

The configuration shown above is not very efficient in transferring energy because only a small portion of the total magnetic flux surrounding the primary conductor will be linked to the secondary circuit. In order to improve the efficiency of the rudimentary transformer, the magnetic field needs to be channeled in such a way that most of the flux produced by the primary conductor is linked to the secondary circuit.

This is accomplished by surrounding the primary and secondary conductors with a magnetic core material having an affinity for magnetic flux. This modification is shown in Figure 1.4. By adding the magnetic core, essentially all of the magnetic flux produced in the primary conductor is linked to the secondary conductor. Therefore, the efficiency of the rudimentary transformer is greatly increased.
Various types of core materials exist. The important physical property is the permeability constant μ, given in units of N/A2. The relative permeability μr is the permeability constant divided by the vacuum permeability μ0.

Values of μr for some common magnetic core materials are as follows:
   SiFe (unoriented) 400
   SiFe (oriented) 1500
   50–50 NiFe (oriented) 2000
   79 Permaloy 12,000–100,000

A grain-oriented silicon steel conducts magnetic flux 1500 times better than a vacuum.  The ratio of the flux density B and the field intensity H is equal to the permeability of the medium μ:
μ = B/H
(1.5.1)
H = B/μ
(1.5.2)

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