Fuji Electric FRENIC-Ace series, User Manual

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Appendix B Japanese Guideline for Suppressing Harmonics by Customers Receiving High
Voltage or Special High Voltage (General-purpose inverter)
Appendix-16
[ 3 ] Examples of calculation
(1) Equivalent capacity
Example of loads
Input capacity and
No. of inverters
Conversion factor Equivalent capacity
[Example (1)] 400V, 3.7kW,10 units
w/- AC/DC reactor
4.61 kVA × 10 units K32=1.4 4.61

10

1.4= 64.54 kVA
[Example (2)] 400V, 1.5kW, 15 units
w/- AC reactor
2.93 kVA × 15 units K34=1.8 2.93

15

1.8= 79.11 kVA
Refer to Table B-2. Refer to Table B-3.
(2) Harmonic current for every harmonic order
Example 1: 400V, 3.7kW, 10 units (w/- AC reactor), maximum availability factor: 0.55
6.6kV side
fundamental current (mA)
Harmonic current onto 6.6 kV lines (mA)
394 × 10= 3940
3940 × 0.55= 2167
5th
(38%)
7th
(14.5%)
11th
(7.4%)
13th
(3.4%)
17th
(3.2%)
19th
(1.9%)
23th
(1.7%)
25th
(1.3%)
823.5 314.2
Refer to Table B-4 and
Table B-6.
Refer to Table B-5.
Example 2: 400V, 3.7kW, 15 units (w/- AC/DC reactor), maximum availability factor: 0.55
6.6kV side
fundamental current (mA)
Harmonic current onto 6.6 kV lines (mA)
394 × 15= 5910
5910 × 0.55= 3250.5
5th
(28%)
7th
(9.1%)
11th
(7.2%)
13th
(4.1%)
17th
(3.2%)
19th
(2.4%)
23th
(1.6%)
25th
(1.4%)
910.1 295.8
Refer to Table B-4 and
Table B-6.
Refer to Table B-5.