Effects of harmonics - Increased losses: Difference between revisions
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== Losses in conductors == | == Losses in conductors == | ||
The active power transmitted to a load is a function of the fundamental component | The active power transmitted to a load is a function of the fundamental component I<sub>1</sub> of the current. | ||
When the current drawn by the load contains harmonics, the rms value of the current, I<sub>r.m.s</sub>, is greater than the fundamental I<sub>1</sub>. | |||
[[Image:Fig_M17.jpg|none]] | |||
'''''Fig.M17:''''' ''Reduced circulation of harmonic currents with detuned reactors'' | |||
'''''Fig. | |||
- | The definition of THD<sub>i</sub> being: | ||
<math>THD_i= \sqrt {\left (\frac{I_{r.m.s}}{I1} \right)^2 - 1}</math> | |||
it may be deduced that : | |||
<math>I_{r.m.s} = I_1 \cdot \sqrt{1 +THD_i ^2}</math> | |||
'''Figure M18''' shows, as a function of the harmonic distortion: | |||
* The increase in the r.m.s. current I<sub>r.m.s.</sub> for a load drawing a given fundamental current | |||
* The increase in Joule losses, not taking into account the skin effect. (The reference point in the graph is 1 for I <sub>r.m.s.</sub> and Joules losses, the case when there are no harmonics) | |||
The harmonic currents cause an increase of the Joule losses in all conductors in which they flow and additional temperature rise in transformers, switchgear, cables, etc. | |||
[[Image:FigM08.jpg|none]] | |||
'''''Fig. M18: '''''<i>Increase in rms current and Joule losses as a function of the THD</i> | |||
== Losses in asynchronous machines == | == Losses in asynchronous machines == | ||
The harmonic voltages (order h) supplied to asynchronous machines | The harmonic voltages (order h) supplied to asynchronous machines cause the flow of currents in the rotor with frequencies higher than 50 Hz that are the origin of additional losses. | ||
=== Orders of magnitude === | |||
* A virtually rectangular supply voltage causes a 20% increase in losses | |||
* A supply voltage with harmonics u<sub>5</sub> = 8% (of U<sub>1</sub>, the fundamental voltage), | |||
u<sub>7</sub> = 5%, u<sub>11</sub> = 3%, u<sub>13</sub> = 1%, i.e. total harmonic distortion THD<sub>u</sub> equal to 10%, results in additional losses of 6% | |||
== Losses in transformers == | |||
Harmonic currents flowing in transformers cause an increase in the “copper” losses due to the Joule effect and increased “iron” losses due to eddy currents. The harmonic voltages are responsible for “iron” losses due to hysteresis. | |||
It is generally considered that losses in windings increase as the square of the THD<sub>i</sub> and that core losses increase linearly with the THD<sub>u</sub>. | |||
In Utility distribution transformers, where distortion levels are limited, losses increase between 10 and 15%. | |||
== Losses in capacitors == | |||
The harmonic voltages applied to capacitors cause the flow of currents proportional to the frequency of the harmonics. These currents cause additional losses. | |||
'''Example''' | |||
A supply voltage has the following harmonics: | |||
* Fundamental voltage U<sub>1</sub> , | |||
* harmonic voltages u<sub>5</sub> = 8% (of U<sub>1</sub>), | |||
* u<sub>7</sub> = 5%, | |||
* u<sub>11</sub> = 3%, | |||
* u<sub>13</sub> = 1%, | |||
i.e. total harmonic distortion THDu equal to 10%. The amperage of the current is multiplied by 1.19. Joule losses are multiplied by (1.19)², i.e. 1.4. | |||
[[ru:Воздействие гармоник -увеличенные потери]] | [[ru:Воздействие гармоник -увеличенные потери]] | ||
[[zh:谐波的影响 - 损耗增大]] | [[zh:谐波的影响 - 损耗增大]] |
Revision as of 18:54, 18 October 2013
Losses in conductors
The active power transmitted to a load is a function of the fundamental component I1 of the current.
When the current drawn by the load contains harmonics, the rms value of the current, Ir.m.s, is greater than the fundamental I1.
Fig.M17: Reduced circulation of harmonic currents with detuned reactors
The definition of THDi being:
[math]\displaystyle{ THD_i= \sqrt {\left (\frac{I_{r.m.s}}{I1} \right)^2 - 1} }[/math]
it may be deduced that :
[math]\displaystyle{ I_{r.m.s} = I_1 \cdot \sqrt{1 +THD_i ^2} }[/math]
Figure M18 shows, as a function of the harmonic distortion:
- The increase in the r.m.s. current Ir.m.s. for a load drawing a given fundamental current
- The increase in Joule losses, not taking into account the skin effect. (The reference point in the graph is 1 for I r.m.s. and Joules losses, the case when there are no harmonics)
The harmonic currents cause an increase of the Joule losses in all conductors in which they flow and additional temperature rise in transformers, switchgear, cables, etc.
Fig. M18: Increase in rms current and Joule losses as a function of the THD
Losses in asynchronous machines
The harmonic voltages (order h) supplied to asynchronous machines cause the flow of currents in the rotor with frequencies higher than 50 Hz that are the origin of additional losses.
Orders of magnitude
- A virtually rectangular supply voltage causes a 20% increase in losses
- A supply voltage with harmonics u5 = 8% (of U1, the fundamental voltage),
u7 = 5%, u11 = 3%, u13 = 1%, i.e. total harmonic distortion THDu equal to 10%, results in additional losses of 6%
Losses in transformers
Harmonic currents flowing in transformers cause an increase in the “copper” losses due to the Joule effect and increased “iron” losses due to eddy currents. The harmonic voltages are responsible for “iron” losses due to hysteresis.
It is generally considered that losses in windings increase as the square of the THDi and that core losses increase linearly with the THDu.
In Utility distribution transformers, where distortion levels are limited, losses increase between 10 and 15%.
Losses in capacitors
The harmonic voltages applied to capacitors cause the flow of currents proportional to the frequency of the harmonics. These currents cause additional losses.
Example
A supply voltage has the following harmonics:
- Fundamental voltage U1 ,
- harmonic voltages u5 = 8% (of U1),
- u7 = 5%,
- u11 = 3%,
- u13 = 1%,
i.e. total harmonic distortion THDu equal to 10%. The amperage of the current is multiplied by 1.19. Joule losses are multiplied by (1.19)², i.e. 1.4.