![Reactive Power (kvar)
2
2
P
S
Q −
=
Active Power
²
² Q
S
P −
=
[KW]
Apparent Power
²
² Q
P
S +
=
[kVA]
cos ϕ = P/S ϕ = phase displacement angle
sin ϕ = Q/S S1 = uncompensated apparent power
Q = S sin ϕ S2 = compensated power with
capacitors for compensation
Q = P tan ϕ
Q1
QC
Q2
ϕ2
ϕ1
S1
S2
Three different types of Power?](https://image.slidesharecdn.com/pfccalculationofharmonicfilters-240518050540-6c5e8b72/85/The-pfc_calculation_of_harmonic_filters-pdf-3-320.jpg)
The pfc_calculation_of_harmonic_filters.pdf
- 1. EPCOS EPCOS EPCOS EPCOS Calculation of harmonic filters EPCOS EPCOS EPCOS EPCOS EPCOS EPCOS EPCOS EPCOS Calculation of harmonic filters Calculation of harmonic filters
- 2. C f U C U Q C f C X X U U Q C C C C C C C C ⋅ ⋅ ⋅ = ⋅ ⋅ = ⋅ ⋅ = ⋅ = ⋅ = π ω π ω 2 2 1 1 2 2 ϕ cos 3 ⋅ ⋅ ⋅ = I U P ϕ sin 3 ⋅ ⋅ ⋅ = I U Q I U S ⋅ ⋅ = 3 S P power Apparent power al factor Power = = _ _ Re _ ( ) 2 1 tan tan ϕ ϕ − ⋅ = P QC Active (Real) power: Reactive power: Apparent power: Required Capacitor output: Basic formulas
- 3. Reactive Power (kvar) 2 2 P S Q − = Active Power ² ² Q S P − = [KW] Apparent Power ² ² Q P S + = [kVA] cos ϕ = P/S ϕ = phase displacement angle sin ϕ = Q/S S1 = uncompensated apparent power Q = S sin ϕ S2 = compensated power with capacitors for compensation Q = P tan ϕ Q1 QC Q2 ϕ2 ϕ1 S1 S2 Three different types of Power?
- 4. C LN R ⋅ = 1 ω The resonant circuit is characterized by a resonant frequency given by: However it is difficult to calculate the value of LN since it depends on the load connected to the network. The resonant frequency can be approximated by the following formula: K 2 R T C u v 100 S Q ⋅ ⋅ < To avoid resonance condition To avoid resonance condition the capacitors output should be the capacitors output should be less than the critical capacitor less than the critical capacitor output calculated by the above output calculated by the above formula: formula: K C T R u Q S f ⋅ ⋅ ⋅ = 100 50 Evaluation of resonance risk
- 5. Example Transformer ST = 630 kVA, uK = 5%, Planned capacitor output QC = 250 kVAr Question: Does the system configuration cause a risk of resonance? According the formula: fR = 50 √ √ √ √ (630*100) / (250*5) = 355 Hz Result: The resonant frequency is close to the 7th harmonic and the capacitor has to be designed for rating below 250 kVAr, or even better, a de-tuned capacitor bank has to be used. K C T R u Q S f ⋅ ⋅ ⋅ = 100 50 Evaluation of resonance risk
- 7. Formulas for calculation of harmonic filters: Reactors connected in series with capacitors result into an increased voltage across the capacitor. Capacitors used for de-tuned filters are therefore required to have voltage ratings higher than the line voltage. p U U N C − ⋅ = 100 100 e.g. UN = 400V, P=7%, calculate UC = 430V A 440V capacitor can be used. C N C C N U U p Q ⋅ ⋅ − = 2 2 100 1 f U p N C N C ⋅ ⋅ ⋅ − ⋅ = π 2 100 1 2 C f p L ⋅ ⋅ ⋅ ⋅ = 2 2 4 100 π 100 2 Re ⋅ = s f f p C L f X X p C L ⋅ ⋅ ⋅ ⋅ ⋅ = ⋅ = 2 2 4 100 100 π De-tuned harmonic filter
- 8. Calculation: example 11 kV level 132 kV level Parallel resonance Iν ν ν ν Iν ν ν ν Series resonance Iν ν ν ν Capacitor bank Transformer 100 kVA, uk = 5 % DC drive 600 kW cosϕ ϕ ϕ ϕ = 0.65 ... 415 V level Transformer 630 kVA, uk = 5 % Capacitor bank 300 kW cosϕ ϕ ϕ ϕ = 0.65 ... 415 V level 3 ˜ if fr = fν ν ν ν Xc 0 Ic ∞ ∞ ∞ ∞
- 10. All components for harmonic filters