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A variable frequency drive datasheet has current parameters named as I2N and I2hd. If Im not mistaken these currents are the per line currents for the drive output as follows where I showed in red color below:

enter image description here

If that is correct the following information makes me confused:

enter image description here

For the yellow encircled VFD nominal power PN above is given as 5.5kW.

But if I use the nominal current and voltage I dont calculate the power as 5.5kW. I calculate as:

PN = sqrt(3)x I x U = 1.73 x 11.9A x 380V = 7.8kW

Is that difference because they are also multiplying with an estimate power factor or is 380V line voltage is wrong? Where am I thinking/knowing wrong?

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  • \$\begingroup\$ The listed power is a mechanical power delivered on an output shaft of induction motor. \$\endgroup\$
    – Marko Buršič
    Commented Mar 27, 2019 at 17:40
  • \$\begingroup\$ But this is from the VFD datasheet there is no motor mentioned. Wouldnt that depend on the motor? I dont get what is meant here if so:( \$\endgroup\$
    – floppy380
    Commented Mar 27, 2019 at 17:47
  • \$\begingroup\$ Yes it is, 5.5Kw or 4kw. Light or heavy duty. \$\endgroup\$
    – Marko Buršič
    Commented Mar 27, 2019 at 17:54

2 Answers 2

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The induction motors come in standard sizes and nominal powers, for example 2.2kW , 3kW, 4kW, 5.5kW, 7.5kW, ...

Those power are the mechanical power, delivered at the output shaft: \$P=M\cdot\omega\$

You can use your inverter with 5.5kW or 4kW induction motor, it depends on the load characteristics. Whenever the load is constant, then it can drvie a 5.5kW motor, but if the load is dynamic with high torque peaks, then a 4kW motor is suitable.

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  • \$\begingroup\$ Thanks and is the I2N and I2hd I marked in red correct ? They are flowing from the drive into the motor correct? The correct meaning of I2N is drive output current ? \$\endgroup\$
    – floppy380
    Commented Mar 27, 2019 at 18:21
  • \$\begingroup\$ Yes, you have correctly marked the output current. \$\endgroup\$
    – Marko Buršič
    Commented Mar 27, 2019 at 18:25
  • \$\begingroup\$ @Marko: I think VFD is better terminology than inverter which can imply a fixed frequency sine-wave inverter which this is not. \$\endgroup\$
    – Transistor
    Commented Mar 27, 2019 at 18:50
  • \$\begingroup\$ @Transistor typically an inverter does not imply a fixed-frequency sine-wave. An inverter is purely DC to AC "invert" the operation of a rectifier (AC - DC) \$\endgroup\$
    – user16222
    Commented Mar 27, 2019 at 21:15
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    \$\begingroup\$ I disagree, this is what I design. Inverter module does not mean a fixed frequency part. It means a 6switch brick \$\endgroup\$
    – user16222
    Commented Mar 27, 2019 at 22:17
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Is that difference because they are also multiplying with an estimate power factor or is 380V line voltage is wrong?

Essentially, both an estimated efficiency and an estimated power factor are assumed. What is more likely is that the full-load current ratings of motors on the market have been surveyed and an effort has been made to accommodate the highest current rating for a given power rating. Some "outlier" data may be neglected. The result is more an estimate of efficiency multiplied by power factor. The end result is to list current ratings that meet the market expectation for a given voltage and power rating.

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