168 AFE200 • Instruction Manual
voutm = 400vca
ηm = 0,96
Im = 43a
Application: continuous load, Pcont= 80%
Requested overload = 150%
> Total electrical power absorbed:
Ptot =
(Poutm * Pcont)
ηm
Ptot =
(18.5 * 0.85)
+
(18.5 * 0.9)
+
(22 * 0.8)
= 52.4 kW
0.95 0.95 0.96
> Total power requested by the DC-link side:
Ptot =
Ptot
+
52.4
= 52.4 kW
ηd 0.97
400 Vac mains, corresponding DC-link value = 650Vcc
Ptot =
Pdc * 1000
+
54000
= 52.4 kW
Vdc-link 650
The AFE200 must be capable of delivering a current of ≥ 83A.
Choose the size with the appropriate characteristics from the section 10.6:
AFE200 - 4450.
> Choosing the drive:
Since the motors are used at below the rated power, the following inverters are
suitable:
Motor 1 : Im1 = 37A * 85% = 31.4A
Choose the size with the rated current required by the motor (≥ 31.4A) from the
catalogue ADV & AFE (section ADV200-DC “2.7 Output Data” on page 37).
With a 650 Vdc drive:
ADV-3185-...-DC (In = 34.2A, OK).
Motor 2 : Im2 = 37A * 90% = 33.3A
Choose the size with the rated current required by the motor (≥ 33.3A) from the
catalogue ADV & AFE (section ADV200-DC “2.7 Output Data” on page 37).
With a 650 Vdc drive:
ADV-3185-...-DC (In = 34.2A, OK).
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