Flux leakage
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Flux Leakage. Main article: Leakage inductance

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Flux Leakage

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Flux leakage

Flux Leakage

Main article: Leakage inductance

The ideal transformer model assumes that all flux generated by the primary winding links all the turns of every winding, including itself. In practice, some flux traverses paths that take it outside the windings.[9] Such flux is termed leakage flux, and manifests itself as self-inductance in series with the mutually coupled transformer windings.[8] Leakage results in energy being alternately stored in and discharged from the magnetic fields with each cycle of the power supply.

http://en.wikipedia.org/wiki/Image:Flux_leakage.svg


Flux leakage1

Flux Leakage

It is not itself directly a source of power loss, but results in poorer voltage regulation, causing the secondary voltage to fail to be directly proportional to the primary, particularly under heavy load. [9] Distribution transformers are therefore normally designed to have very low leakage inductance.


Flux leakage exploited

Flux Leakage…. exploited

  • However, in some applications, leakage can be a desirable property, and long magnetic paths, air gaps, or magnetic bypass shunts may be deliberately introduced to a transformer's design to limit the short-circuit current it will supply. [8] Leaky transformers may be used to supply loads that exhibit negative resistance, such as electric arcs, mercury vapor lamps, and neon signs; or for safely handling loads that become periodically short-circuited such as electric arc welders. [10] Air gaps are also used to keep a transformer from saturating, especially audio-frequency transformers that have a DC component added.


Effect of frequency

Effect of frequency

  • In practice, the flux would rise very rapidly to the point where magnetic saturation of the core occurred, causing a huge increase in the magnetizing current and overheating the transformer. All practical transformers must therefore operate under alternating (or pulsed) current conditions. [12]


Effect of frequency1

Effect of frequency

  • The EMF of a transformer at a given flux density increases with frequency, an effect predicted by the universal transformer EMF equation.[6] By operating at higher frequencies, transformers can be physically more compact because a given core is able to transfer more power without reaching saturation, and fewer turns are needed to achieve the same impedance. However properties such as core loss and conductor skin effect also increase with frequency.


Emf in transformers

EMF in Transformers

Transformer universal EMF equation

  • If the flux in the core is sinusoidal, the relationship for either winding between its rms EMF E, and the supply frequency f, number of turns N, core cross-sectional area a and peak magnetic flux densityB is given by the universal EMF equation: [6]


Emf in transformers1

EMF in Transformers

  • Aircraft and military equipment traditionally employ 400 Hz power supplies which are less efficient but this is more than offset by the reduction in core and winding weight. [13]


At higher frequency

At Higher frequency

  • In general, operation of a transformer at its designed voltage but at a higher frequency than intended will lead to reduced magnetising current. At a frequency lower than the design value, with the rated voltage applied, the magnetising current may increase to an excessive level.


At higher frequency1

At Higher frequency

  • Operation of a transformer at other than its design frequency may require assessment of voltages, losses, and cooling to establish if safe operation is practical. For example, transformers may need to be equipped with "volts per hertz" over-excitation relays to protect the transformer from overvoltage at higher than rated frequency.


Energy losses

Energy losses

  • An ideal transformer would have no energy losses, and would therefore be 100% efficient. Despite the transformer being amongst the most efficient of electrical machines, with experimental models using superconducting windings achieving efficiencies of 99.85%, [14] energy is dissipated in the windings, core, and surrounding structures. Larger transformers are generally more efficient, and those rated for electricity distribution usually perform better than 95%. [15] A small transformer, such as a plug-in "power brick" used for low-power consumer electronics, may be no more than 85% efficient; although individual power loss is small, the aggregate losses from the very large number of such devices is coming under increased scrutiny. [16]


Energy losses1

Energy losses

  • Transformer losses are attributable to several causes and may be differentiated between those originating in the windings, sometimes termed copper loss, and those arising from the magnetic circuit, sometimes termed iron loss. The losses vary with load current, and may furthermore be expressed as "no-load" or "full-load" loss, respectively. Winding resistance dominates load losses, whereas hysteresis and eddy currents losses contribute to over 99% of the no-load loss. The no-load loss can be significant, meaning that even an idle transformer constitutes a drain on an electrical supply, and lending impetus to development of low-loss transformers (also see energy efficient transformer). [17]


Examples

Examples


Example

Example

  • The transformer is the most common application of the concept of mutual inductance. In the transformer, the effect of the mutual inductance is to cause the primary ciruit to take more power from the electrical supply in response to an increased load on the secondary. For example, if the load resistance in the secondary is reduced, then the power required will increase, forcing the primary side of the transformer to draw more current to supply the additional need.


Circuit equations

Circuit Equations


Power in transformer

Power in Transformer


Power in transformer1

Power in Transformer


Power in transformer2

Power in Transformer


Example1

Example

  • A 200 KVA single phase transformer with a voltage ratio 6350/660 V has the following windings resistance and reactance:

    R1=1.56k; R2=0.061; x1=4.67; x2=0.048

    No load I=0.96A at p.f. of 0.263. Calculate equivalent circuit parameters referred to High voltage side.


Solution

Solution

Re=R1+R2(N1/N2)2

xe=x1+x2(N1/N2)2


Vector analysis

Vector Analysis

  • So, the first question is: how do I know which phase I am measuring?


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