Guide for electrical design engineers
Power Po wer Quali Qualitty Krzysztof Piatek AGH-University of Science & Technology
Centralised reactive power compensation in case of harmonic pollution
I ( n)
n X tr
X c n
P o w e r Q u a l
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Problem A 400V switchboard is ed rom the transormer with rated power SN=400kV =400kVA A and short-circuit shor t-circuit voltage e%=4.5%. The switchboard supplies loads with total active power ranging rom P=190 kW to 208kW and a power actor varying rom 0.62 to 0.68. The load current is distorted; maximum values o harmonic currents are given in the table below. n
5
7
11
13
I(n) [A]
10
7.14
5.13
4.1
Determine the maximum and minimum compensator power in order to compensate reactive power to the power actor value o 0.92. Check the voltage distortion is maintained within allowed limits and calculate the series reactor impedance, i needed. For the purpose o this calculation the power system short-circuit capacity, resistances o components and compensator active power losses can be neglected. Assume the supply voltage is non-distorted.
Reactive power compensation The reactive power needed to compensate a load can be calculated rom the reactive power balance. As the active power does not change in result o compensation, the ollowing ormula can be used Qk = P ( tan ϕn − tan ϕd )
where: Qk – reactive power necessary or compensation; P – the load active power; tanϕn – non-compensated tangent value; tanϕd – tangent value that should be achieved by the compensation. As the active power varies also the power actor and the load reactive power are variable. Since there is no inormation on the correlation between the power actor and the active power, power, we shall assume extreme cases and calculate both the maximum and minimum compensating power or the given tangent value. The maximum power is drawn at the maximum active power and the maximum diference between tangent values, i.e. at the minimum power actor o 0.62. Correspondingly, the minimum power occurs or the minimum active power and maximum power actor 0.68. Thereore, we obtain Qk ,min =190 (1.07 078 − 0. 42 426) = 123. 93 93 kVAr Qk ,max = 208 (1.26 265 − 0.42 4 26) =174 1 74.61 kVAr
Thus the compensator reactive power has a certain constant value o Qst = 123,93 kVAr, whereas the regulation range is Qreg = 50,61 kVAr. Since the load current is distorted, there is a possibility a resonance may occur. Moreover, neglecting the resistance implies that attenuation introduced by certain elements will not be taken into account. It should be noted that harmonic currents values are small compared to the 400 kVA transormer transormer rated current. For high harmonic requencies the transormer, capacitor bank and the load orm a parallel circuit as shown in gure below.
I ( n)
n X tr
X c n
Where: I(n) – n-th harmonic source current; Xtr – the transormer reactance calculated or the rst harmonic; Xc – the compensator reactance the rst harmonic; n – relative requency (with respect to the undamental — a harmonic order).
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In this case the equivalent impedance (actually the reactance, assuming the resistance is neglected) is −nX X Z z (n) = 2 c tr n X tr − X c The resonance conditions are related to the zeros o the numerator or denominator. For positive values o circuit components only the numerator can become zero. This occurs or n2 X tr = X c and is the condition o a parallel resonance; the resonance requency is nr =
X c X tr
The reactance values can be determined rom simple ormulas: X c ,min = X c ,max =
U 2 Qk ,max U 2 Qk ,max
=
400 14761 . ⋅103
=
= 0.916 916 Ω
400 123.93⋅103
=1.291 291 Ω
e% U 2 400 = 0.045 =18 mΩ X tr = 045⋅ 100 Str 400⋅103 As the compensator reactive power varies also its reactance does and, consequently, the resonance requency is varying. Substituting the above values we determine the maximum and minimum resonance requency: •
nr, max = 8.47
or Qr, min, Xr, max
•
nr, min = 7.13
or Q r, max, X r, min
Since the 7th harmonic is present in the load current, a resonance can occur when the capacitor bank is loaded with maximum power. This will result in an increase in the 7th harmonic current and, consequently, signifcant voltage distortion at the point p oint o common coupling (at the switchboard bus-bars). In order to check whether the capacitor bank operation is possible under such conditions, the maximum total harmonic voltage distortion THD should be calculated. The resonance will obviously occur at the requency nr equal 7.13, i.e. or the capacitor maximum reactive power. Knowing maximum harmonic current values, the voltage drops and hence the voltage THD can be calculated. The table below shows the values calculated or each harmonic. n
5
7
11
13
I(n) [A]
10
7.14
5.13
4.1
Z(n) [Ω]
0.177
3.365
0.144
0.1
U(n) [ V ]
1.768
24.03
0.738
0.414
Total T otal harmonic voltage distortion THD can be determined rom these values according to the defnition
THD =
∑U
2 ( n)
U N (1)
⋅100%
where UN(1) – rms value o the phase voltage (the frst harmonic); U(n) – rms value o the n-th harmonic voltage drop. The calculated value o total harmonic distortion THD will be 1.7682 + 24.032 + 0.7082 + 0. 4142 ⋅100% =10.48% THD = 230 This value exceeds the limit o 8% set orth in standard EN 50160 with regard to low voltage. Such high a voltage distortion occurs only at the maximum capacitor bank power, thereore either the power needs to be reduced or a series reactor should be connected.
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Voltage Vo ltage harmonics suppression Connecting a series reactor creates conditions or series resonance. The equivalent circuit or this case is shown in fgure below.
I (n)
nXd n X tr
Xc n
Where Xd – is the reactance o the series reactor. The equivalent impedance o this circuit is Z z (n) =
nX tr (n2 X d − X c ) 2
n
( X tr + X d )− X c
Two options are possible when the numerator or denominator equals zero. The nominator becomes zero i the relationship n2 X d − X c = 0 is satisfed, which is the condition o series resonance. I the resonance requency is known the reactor reactance can be calculated rom X d =
X c ns2
For the denominator we obtain the relationship n2 ( X tr + X d )− X c = 0 which is the condition o parallel resonance. The resonance requency is nr =
X c X tr + X d
It should be noted that the parallel resonance requency has been shited toward lower requencies compared to that o the circuit without the series reactor. Thereore, Thereore, when choosing the reactor the series resonance requency should be selected to match the lowest harmonic present — in this case the 5th. Choosing the series resonance requency or e.g. the 7th harmonic will shit shi t the parallel resonance requency towards the 5th harmonic. This may increase the voltage distortion compared to t hat o the circuit without the reactor. As an example, choosing the reactor or the 7th harmonic and using the values given in this problem, we obtain X d =
X c ns2
=
0.919 919 72
=18.55 mΩ
But the parallel resonance will occur or the requency nr =
X c X tr + X d
= 4.99
and, in consequence, the voltage distortion THD will be approximately 146%.
Centralised reactive power compensation in case of harmonic pollution http://www.leonardo-energy.org
Thereore the reactor shall be chosen or the requency equal to the 5th harmonic. Then we obtain X d =
X c ns2
=
0.919 919 52
= 3676 . mΩ
The parallel resonance requency will vary within the range nr ,max = 4.86 nr ,min = 4.09 It is thus sufciently ar rom the current harmonics h armonics (namely the 5th harmonic). Now we calculate the maximum voltage distortion THD which will occur or the minimum compensator power, i.e. at the resonant requency 4.86. The calculations are tabulated below: n
5
7
11
13
I(n)
10
7.14
5.13
4.1
Z(n)
0.448
0.046
0.117
0.144
U(n)
4.48
0.327
0.6
0.592
Total T otal voltage distortion THD is 4.482 + 0.3272 + 0.62 + 0. 5922 ⋅100% =1.99% 230
THD =
which is ar below the admissible value. The graph o the modulus o impedance versus requency is shown in gure 1b. The correct selection o the compensator components, i.e. the capacitors and reactor, requires determining their operating conditions. The current in the compensator branch can be ound employing Kirchho current law. law. If ( n) = I ( n)
n2 X tr n2 ( X tr + X d )− X c
where: I(n) – rms n-th harmonic current, Xd, Xtr, Xc – components reactances or the undamental harmonic. The undamental harmonic current we calculate rom the ormula: I f (1) =
U 3 ( X c − X d )
= 262 262.53 A
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The worst-case calculation results are tabulated below n
5
7
11
13
I(n)
10
7.14
5.13
4.1
I(n)
59.81
4.54
2.1
1.57
The rms current value we calculate rom the ormula:
∑I
2 ( n)
RMS = This yields:
RMS = 262.532 + 59.812 + 4.542 + 2. 12 +1. 572 = 269. 31 A This value should be smaller than the admissible current value Irms =1.8⋅I CN
where ICN is the rated capacitor current. The capacitor voltage increase can be ound rom the Kirchho voltage law, or using the calculated values o the compensator branch current. Thus we obtain: n
5
7
11
13
I(n)
59.81
4.54
2.1
1.57
XC(n)
0.183
0.131
0.083
0.07
UC(n)
10.94
0.59
0.17
0.11
From the voltage divider the capacitor undamental harmonic voltage rms value is U C (1) =
X c X c − X d
U = 416 416.67 V
which yields the rms value: RMS = 240.562 +10.942 + 0.59 2 + 0.172 + 0. 112 = 240. 81 V This value should be less than the maximum permissible continuous voltage, determined as UC ,rms <1.2U CN
where UCN – the rated capacitor voltage. The peak capacitor voltage is determined rom the ormula MAX =
∑U
( n)
This yields MAX = 240.56 +10.94 + 0.59 + 0.17 + 0. 11= 252. 37 37 V This value should be less than the maximum permissible peak voltage, which is UC ,max <1.1U CN Using these parameters we can speciy the size and type o capacitors and reactors appropriate or the given compensator requirements.
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