Voltage and Current Harmonics (John Wiley & Sons)

Author: Angelo Baggini and Zbigniew Hanzelka

Source: Handbook of Power Quality Edited by Angelo Baggin, John Wiley & Sons, Ltd

1.0 SELECTION AND RATING OF TRANSFORMERS FOR A SIX-PULSE CONVERTER [10]

When the harmonic spectrum is known, or at least can be measured with a certain reliability or predicted, the additional losses can be easily calculated.

The process of calculation should be made through the following steps:

  1. Determination of all the coponents of additional losses due to the presence of harmonics.
  2. Determination of the harmonic spectrum, either by measurement or by estimation, taking into account all harmonic generating equipment, in particular electronic converters.
  3. Calculation of the contribution of each harmonic component and determination of total additional losses.

In practice, it is important to use the real harmonic current magnitudes rather than theoretical values.

Table  1 shows the calculated additional losses, for harmonic currents up to order 25, for two transformers at normal environmental temperature, assuming the current harmonic spectrum illustrated in Figure 1.

Table 1 Additional losses calculated in the presence of non-sinusoidal

Table 1 Additional losses calculated in the presence of non-sinusoidal

Figure 1 Theoretical and actual values of current harmonics for a six-pulse converter (in pu)

Figure 1 Theoretical and actual values of current harmonics for a six-pulse converter (in pu)

The results demonstrate that the transformer characteristics play an important role in determining the losses with harmonic loads.

The transformers in this example were measured at slightly different temperatures (21.5°C for the first and 22.8°C for the second); this will not change the reliability of results.

 1.1 Calculation of the K Factor

Table 2 shows the calculation of the K factor for the harmonic spectrum of Figure  1 on a per unit basis.

The first step is the calculation of the r.m.s. value of total current I, 1.0410 in this case, after which the squares of the proportionate values of each harmonic current can be calculated, leading to the value of K. For such a load, a transformer with a K rating of 9 would be appropriate for a six-pulse converter.

Table 2 Reduction factors for current harmonics

Table 2 Reduction factors for current harmonics

1.2 Calculation of the Factor K

The first step in establishing factor K (Table 2) is to discover the value of e, the ratio of eddy current loss to total load loss at fundamental frequency. The transformer manufacturer should be able to provide this, otherwise it is likely to lie in the range of 0.05 to 0.1. The exponent q depends critically on the construction of the transformer and should also be available from the manufacturer. It is likely to lie in the range 1.5 to 1.7. As before, the calculations are based on the theoretical values from Figure 1. In practice, the transformer would need to be derated to 84.75 % (1/1.18) of nominal power rating when supplying a six-pulse converter.

Table 3 Reduction factors for current harmonics

Table 3 Reduction factors for current harmonics

2.0 DERATING CABLES

As described in Section 6.2, the current amplitude in the neutral due to the third harmonic could exceed in amplitude the phase current at the fundamental frequency. In this case the neutral current should be considered with regard to the sizing of the circuit cables. This example is related to an office building where four different harmonics spectra have been used to evaluate the cable size to be installed.

The system is a three-phase circuit with a 32 A rated load to be installed using a four-core EPR insulated cable laid directly onto the wall.

2.1 Scenarios

These are as follows:

  1. Absence of harmonics. For this current it is common practice to use a copper conductor cable with a 4 mm2 cross-section with a capacity of 35 A [5] .
  2. A value of 22 % of the third-order harmonic (Figure  2). For this spectrum the neutral current will be IN = 32·0,22·3 = 21,1A, IN <IF, so the value is selected on the basis of the line current. Applying a 0.86 reduction factor (Table 12), the equivalent load current is 32/0,86 =37,2 A. For this value the cable section hasa6mm2 cross-section with a capacity of 44 A [5].

For a value of 42 % of the third-order harmonic (Figure 3), IN = 32·0,42·3=40.3A, IN >IF, so the value is selected on the basis of the neutral current. Applying a 0.86 reduction factor, the equivalent load current is 40,3/0,86=46,9 A. For this value the cable section has a 10 mm2 cross-section with a capacity of 60A [5].

Figure 2 Current waveform and its spectrum

Figure 2 Current waveform and its spectrum

Figure 3 Current waveform and its spectrum

Figure 3 Current waveform and its spectrum

Figure 4 Current waveform and its spectrum

Figure 4 Current waveform and its spectrum

Figure C7.4 Current waveform and its spectrum

3. Third-order, harmonic-rich environment, as in Figure 4. The neutral current will be In= 32·1.31·3 = 125.76A, In>If, so the value is selected on the basis of the neutral current. Applying a reduction factor equal to 1, the equivalent load current is 125.76/1 = 125,67 A. For this value the cable section has a 35 mm2 cross-section with a capacity of 128 A [5].

3.0 HARMONIC SOURCE LOCATION

In the event of significant distortion of the supply network voltage at the PCC between the electricity supplier and customer, the source of disturbance should be located. This becomes of particular significance when formulating contracts for electric power supply or charging for worsening the quality of supply.Inmanycasesalsoaquantitative determination of the supplier and customer(s) contribution to the total voltage distortion at the PCC is required.

Figure 5 The principle of locating the nth harmonic source on the basis of its active power measurement

Figure 5 The principle of locating the nth harmonic source on the basis of its active power measurement

The most common practical method for locating harmonic sources is based on deter­mining the direction of active power flow for given harmonics, though many authors indicate its limitations and propose others methods (investigation of the direction of reac­tive power flow and the ‘critical impedance’, interharmonic injection, determining voltage and current relative values, etc. [34],[35]). In most cases these methods, apart from their technical complexity, require precise information on values of equivalent parameters of the analyzed system, which are difficult to access, or can only be obtained as a result of costly measurements.

According to the direction of active power flow method, the dominant source of a given harmonic (of order n) can be located by determining the direction of this harmonic active power flow at various points of the system (Figure 5). A non-zero value of P(n) =U(n)I(n) cos(Φiu(n)−Φii(n))is the effect of the interaction of voltage and current with the same frequency. A linear load supplied with distorted voltage draws active power for each harmonic: P(n) ≥ 0. If non-linear elements exist at the customer side, the active power for someharmonicscan besuppliedtothenetwork: P(n)<0. Thesignof P(n) canbedetermined by means of measuring the phase angles of the voltage and current of the same order: Φiu(n)and Φi(n).

The principle of this method is explained in the example of a single-phase circuit, shown in Table 4 (the supply voltage source is US, LS), where the non­linear load is the thyristor power controller (TYR1, TYR2, resistance RONL, induc­tance LONL), which is the source of harmonic currents of order n = 2k ± 1 = 3, 5, 7, 9, 11, 13, 15, (for k = 1,2,3,…). There cases, distinguished by location of the voltage distortion source, are discussed for the power controller located: (i) upstream of the PCC, (ii) downstream of the PCC, and (iii) harmonic sources at both sides of the PCC

Voltage and Current Harmonics 9

Voltage and Current Harmonics 10

Table 4 Example simulations illustrating the method for harmonic source location based on the active power measurement

 

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