Fuji Electric FRENIC MEGA G2 Series - 3 ] Examples of Calculation

Appendix B Japanese Guideline for Suppressing Harmonics by Customers Receiving High
Voltage or Special High Voltage (General-purpose Inverters)
Appendix-18
[ 3 ] Examples of calculation
(1) Equivalent capacity
Example of loads
Input capacity and
No. of inverters
Conversion factor
Equivalent capacity
[Example (1)] 400 V, 3.7 kW, 10 units
with AC/DC reactor
4.61 kVA x 10 units
K32 = 1.4
4.61 x 10 x 1.4 = 64.54 kVA
[Example (2)] 400 V, 1.5 kW, 15 units
with AC reactor
2.93 kVA x 15 units
K34 = 1.8
2.93 x 15 x 1.8 = 79.11 kVA
See Table B.2-1
See Table B.2-2
(2) Harmonic current for every harmonic order
Example 1: 400 V, 3.7 kW, 10 units (with AC reactor), maximum availability factor: 0.55
6.6 kV side
fundamental current (mA)
Harmonic current onto 6.6 kV lines (mA)
394 x 10 = 3940
3940 x 0.55 = 2167
5th
(38%)
7th
(14.5%)
11th
(7.4%)
13th
(3.4%)
17th
(3.2%)
19th
(1.9%)
23rd
(1.7%)
25th
(1.3%)
823.5
314.2
See Table B.2-3 and
Table B.2-5
See Table B.2-4
Example 2: 400 V, 3.7 kW, 15 units (with AC/DC reactor), maximum availability factor: 0.55
6.6 kV side
fundamental current (mA)
Harmonic current onto 6.6 kV lines (mA)
394 x 15 = 5910
5910 x 0.55 = 3250.5
5th
(28%)
7th
(9.1%)
11th
(7.2%)
13th
(4.1%)
17th
(3.2%)
19th
(2.4%)
23rd
(1.6%)
25th
(1.4%)
910.1
295.8
See Table B.2-3 and Table
B.2-5
See Table B.2-4