Based on the installation I want to calculate the acceptable current harmonics related to my installation. IEC 61000-3-6 says:

For each customer only a fraction of the global emission limits GhMV+LV will be allowed. A reasonable approach is to take the ratio between the agreed power Si and the total supply capability St of the MV system. Such a criterion is related to the fact that the agreed power of a customer is often linked with his share in the investment costs of the power system.


\$EUhi = Gh_{MV+LV} \cdot \sqrt[\alpha]{\frac{S_i}{S_t}}\$

EUhi is the harmonic voltage emission limit of order h for the installation (i) directly supplied at MV (%),

GhMV+LV is the maximum global contribution of the total of MV and LV installations that can be supplied from the considered MV system to the hth harmonic voltage in the MV system,

Si = Pi /cosϕi is the agreed power of customer installation i, or the MVA rating of the considered distorting installation (either load or generation), St is the total supply capacity of the considered system including provision for future load growth,

St is the sum of the capacity allocations of all installations including that of downstream installations that are or can be connected to the considered system, taking diversity into consideration). St might also include the contribution from dispersed generation, however more detailed consideration will be required to determine its firm contribution to St and its effective contribution to the short-circuit power as well,

α is the summation law exponent (see Table).

Finally, they divide EUhi by the grid harmonics impedance Zhi (a property of the grid, to my understanding independent from my installation) in order to get the acceptable current harmonics for the customer installation. An example of the calculation is given e.g. in Appendix A of this paper.

I have problems to understand the following issue which I am working on:

Let's assume I have a 10 kV distribution with capacity St=20 MW, and I would install two converters of Si = 2 MW each. One converter's acceptable 5th harmonic would be e.g. 5% which would be 0.05 * 160 A = 8 A. So both converters would together contribute 16 A of a 5th harmonic.

If I would, alternatively, install a single converter of Si = 4 MW, according to the above equation, EUhi would increase (e.g. from \$\sqrt[1.4]{\frac{2 MW}{20 MW}} = 0.193\$ to \$\sqrt[1.4]{\frac{4 MW}{20 MW}} = 0.317\$ for alpha = 1.4), and the 4 MW converter would then be allowed of contribute 1.64 higher current harmonics at the 5th harmonic, which would be 0.05 * 1.64 * 320 A = 26 A.

Is my interpretation of IEC 61000-3-6 wrong? Do I misunderstand something? Or is it really like that? If it is correct, how does it make sense?



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