Miscalculating the nominal power using nominal current and voltage of a VFDCan we use a power amplifier (Class AB) in the inverter portion of a VFD?Startup current of an induction motor when using VFDVFD Power Savings physical explanation - cubic vs square relationshipI too want to wire a 3 phase motor to a VFD. And the question has already been asked butTesting an induction motor with a VFD, difference in voltage and mechanical power vs electric powerIdentifying a multiple secondary unknown transformerCan I use a VFD to send power of a specific frequency to a frequency meter?Power Distribution Units and Nominal Input VoltageVFD? to power aircraft non-motor/inductive load OR power conversionHow to measure output voltage and current of a VFD?
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Miscalculating the nominal power using nominal current and voltage of a VFD
Can we use a power amplifier (Class AB) in the inverter portion of a VFD?Startup current of an induction motor when using VFDVFD Power Savings physical explanation - cubic vs square relationshipI too want to wire a 3 phase motor to a VFD. And the question has already been asked butTesting an induction motor with a VFD, difference in voltage and mechanical power vs electric powerIdentifying a multiple secondary unknown transformerCan I use a VFD to send power of a specific frequency to a frequency meter?Power Distribution Units and Nominal Input VoltageVFD? to power aircraft non-motor/inductive load OR power conversionHow to measure output voltage and current of a VFD?
$begingroup$
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:
If that is correct the following information makes me confused:
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?
power vfd
asked 5 hours ago
HelpMeeHelpMee
4901029
$endgroup$
add a comment |
$begingroup$
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:
If that is correct the following information makes me confused:
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?
power vfd
asked 5 hours ago
HelpMeeHelpMee
4901029
$endgroup$
$begingroup$
The listed power is a mechanical power delivered on an output shaft of induction motor.
$endgroup$
– Marko Buršič
5 hours ago
$begingroup$
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:(
$endgroup$
– HelpMee
5 hours ago
$begingroup$
Yes it is, 5.5Kw or 4kw. Light or heavy duty.
$endgroup$
– Marko Buršič
4 hours ago
add a comment |
$begingroup$
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:
If that is correct the following information makes me confused:
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?
power vfd
asked 5 hours ago
HelpMeeHelpMee
4901029
$endgroup$
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:
If that is correct the following information makes me confused:
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?
power vfd
power vfd
asked 5 hours ago
HelpMeeHelpMee
4901029
asked 5 hours ago
HelpMeeHelpMee
4901029
asked 5 hours ago
HelpMeeHelpMee
4901029
asked 5 hours ago
HelpMeeHelpMee
4901029
asked 5 hours ago
HelpMeeHelpMee
4901029
4901029
$begingroup$
The listed power is a mechanical power delivered on an output shaft of induction motor.
$endgroup$
– Marko Buršič
5 hours ago
$begingroup$
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:(
$endgroup$
– HelpMee
5 hours ago
$begingroup$
Yes it is, 5.5Kw or 4kw. Light or heavy duty.
$endgroup$
– Marko Buršič
4 hours ago
add a comment |
$begingroup$
The listed power is a mechanical power delivered on an output shaft of induction motor.
$endgroup$
– Marko Buršič
5 hours ago
$begingroup$
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:(
$endgroup$
– HelpMee
5 hours ago
$begingroup$
Yes it is, 5.5Kw or 4kw. Light or heavy duty.
$endgroup$
– Marko Buršič
4 hours ago
$begingroup$
The listed power is a mechanical power delivered on an output shaft of induction motor.
$endgroup$
– Marko Buršič
5 hours ago
$begingroup$
The listed power is a mechanical power delivered on an output shaft of induction motor.
$endgroup$
– Marko Buršič
5 hours ago
$begingroup$
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:(
$endgroup$
– HelpMee
5 hours ago
$begingroup$
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:(
$endgroup$
– HelpMee
5 hours ago
$begingroup$
Yes it is, 5.5Kw or 4kw. Light or heavy duty.
$endgroup$
– Marko Buršič
4 hours ago
$begingroup$
Yes it is, 5.5Kw or 4kw. Light or heavy duty.
$endgroup$
– Marko Buršič
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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: $P=Mcdotomega$
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.
answered 4 hours ago
Marko BuršičMarko Buršič
10.3k2812
$endgroup$
$begingroup$
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 ?
$endgroup$
– HelpMee
4 hours ago
$begingroup$
Yes, you have correctly marked the output current.
$endgroup$
– Marko Buršič
4 hours ago
$begingroup$
@Marko: I think VFD is better terminology than inverter which can imply a fixed frequency sine-wave inverter which this is not.
$endgroup$
– Transistor
4 hours ago
$begingroup$
@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)
$endgroup$
– JonRB
1 hour ago
$begingroup$
@JonRB: I know. I use both VFDs and sinewave inverters in industrial applications but when purchasing sinewave inverters for 60 Hz equipment for use in Europe vendors, who should know better, assure me with confidence that a VFD is what I need. (It's not.) I think the distinction is worth making.
$endgroup$
– Transistor
1 hour ago
|
show 1 more comment
$begingroup$
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.
answered 2 hours ago
Charles CowieCharles Cowie
21.3k11639
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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: $P=Mcdotomega$
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.
answered 4 hours ago
Marko BuršičMarko Buršič
10.3k2812
$endgroup$
$begingroup$
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 ?
$endgroup$
– HelpMee
4 hours ago
$begingroup$
Yes, you have correctly marked the output current.
$endgroup$
– Marko Buršič
4 hours ago
$begingroup$
@Marko: I think VFD is better terminology than inverter which can imply a fixed frequency sine-wave inverter which this is not.
$endgroup$
– Transistor
4 hours ago
$begingroup$
@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)
$endgroup$
– JonRB
1 hour ago
$begingroup$
@JonRB: I know. I use both VFDs and sinewave inverters in industrial applications but when purchasing sinewave inverters for 60 Hz equipment for use in Europe vendors, who should know better, assure me with confidence that a VFD is what I need. (It's not.) I think the distinction is worth making.
$endgroup$
– Transistor
1 hour ago
|
show 1 more comment
$begingroup$
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: $P=Mcdotomega$
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.
answered 4 hours ago
Marko BuršičMarko Buršič
10.3k2812
$endgroup$
$begingroup$
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 ?
$endgroup$
– HelpMee
4 hours ago
$begingroup$
Yes, you have correctly marked the output current.
$endgroup$
– Marko Buršič
4 hours ago
$begingroup$
@Marko: I think VFD is better terminology than inverter which can imply a fixed frequency sine-wave inverter which this is not.
$endgroup$
– Transistor
4 hours ago
$begingroup$
@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)
$endgroup$
– JonRB
1 hour ago
$begingroup$
@JonRB: I know. I use both VFDs and sinewave inverters in industrial applications but when purchasing sinewave inverters for 60 Hz equipment for use in Europe vendors, who should know better, assure me with confidence that a VFD is what I need. (It's not.) I think the distinction is worth making.
$endgroup$
– Transistor
1 hour ago
|
show 1 more comment
$begingroup$
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: $P=Mcdotomega$
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.
answered 4 hours ago
Marko BuršičMarko Buršič
10.3k2812
$endgroup$
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: $P=Mcdotomega$
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.
answered 4 hours ago
Marko BuršičMarko Buršič
10.3k2812
answered 4 hours ago
Marko BuršičMarko Buršič
10.3k2812
answered 4 hours ago
Marko BuršičMarko Buršič
10.3k2812
answered 4 hours ago
Marko BuršičMarko Buršič
10.3k2812
10.3k2812
$begingroup$
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 ?
$endgroup$
– HelpMee
4 hours ago
$begingroup$
Yes, you have correctly marked the output current.
$endgroup$
– Marko Buršič
4 hours ago
$begingroup$
@Marko: I think VFD is better terminology than inverter which can imply a fixed frequency sine-wave inverter which this is not.
$endgroup$
– Transistor
4 hours ago
$begingroup$
@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)
$endgroup$
– JonRB
1 hour ago
$begingroup$
@JonRB: I know. I use both VFDs and sinewave inverters in industrial applications but when purchasing sinewave inverters for 60 Hz equipment for use in Europe vendors, who should know better, assure me with confidence that a VFD is what I need. (It's not.) I think the distinction is worth making.
$endgroup$
– Transistor
1 hour ago
|
show 1 more comment
$begingroup$
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 ?
$endgroup$
– HelpMee
4 hours ago
$begingroup$
Yes, you have correctly marked the output current.
$endgroup$
– Marko Buršič
4 hours ago
$begingroup$
@Marko: I think VFD is better terminology than inverter which can imply a fixed frequency sine-wave inverter which this is not.
$endgroup$
– Transistor
4 hours ago
$begingroup$
@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)
$endgroup$
– JonRB
1 hour ago
$begingroup$
@JonRB: I know. I use both VFDs and sinewave inverters in industrial applications but when purchasing sinewave inverters for 60 Hz equipment for use in Europe vendors, who should know better, assure me with confidence that a VFD is what I need. (It's not.) I think the distinction is worth making.
$endgroup$
– Transistor
1 hour ago
$begingroup$
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 ?
$endgroup$
– HelpMee
4 hours ago
$begingroup$
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 ?
$endgroup$
– HelpMee
4 hours ago
$begingroup$
Yes, you have correctly marked the output current.
$endgroup$
– Marko Buršič
4 hours ago
$begingroup$
Yes, you have correctly marked the output current.
$endgroup$
– Marko Buršič
4 hours ago
$begingroup$
@Marko: I think VFD is better terminology than inverter which can imply a fixed frequency sine-wave inverter which this is not.
$endgroup$
– Transistor
4 hours ago
$begingroup$
@Marko: I think VFD is better terminology than inverter which can imply a fixed frequency sine-wave inverter which this is not.
$endgroup$
– Transistor
4 hours ago
$begingroup$
@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)
$endgroup$
– JonRB
1 hour ago
$begingroup$
@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)
$endgroup$
– JonRB
1 hour ago
$begingroup$
@JonRB: I know. I use both VFDs and sinewave inverters in industrial applications but when purchasing sinewave inverters for 60 Hz equipment for use in Europe vendors, who should know better, assure me with confidence that a VFD is what I need. (It's not.) I think the distinction is worth making.
$endgroup$
– Transistor
1 hour ago
$begingroup$
@JonRB: I know. I use both VFDs and sinewave inverters in industrial applications but when purchasing sinewave inverters for 60 Hz equipment for use in Europe vendors, who should know better, assure me with confidence that a VFD is what I need. (It's not.) I think the distinction is worth making.
$endgroup$
– Transistor
1 hour ago
|
show 1 more comment
$begingroup$
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.
answered 2 hours ago
Charles CowieCharles Cowie
21.3k11639
$endgroup$
add a comment |
$begingroup$
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.
answered 2 hours ago
Charles CowieCharles Cowie
21.3k11639
$endgroup$
add a comment |
$begingroup$
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.
answered 2 hours ago
Charles CowieCharles Cowie
21.3k11639
$endgroup$
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.
answered 2 hours ago
Charles CowieCharles Cowie
21.3k11639
answered 2 hours ago
Charles CowieCharles Cowie
21.3k11639
answered 2 hours ago
Charles CowieCharles Cowie
21.3k11639
answered 2 hours ago
Charles CowieCharles Cowie
21.3k11639
21.3k11639
add a comment |
add a comment |
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$begingroup$
The listed power is a mechanical power delivered on an output shaft of induction motor.
$endgroup$
– Marko Buršič
5 hours ago
$begingroup$
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:(
$endgroup$
– HelpMee
5 hours ago
$begingroup$
Yes it is, 5.5Kw or 4kw. Light or heavy duty.
$endgroup$
– Marko Buršič
4 hours ago